473 lines
14 KiB
Markdown
473 lines
14 KiB
Markdown
<p align="center">
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<a href="https://www.programmercarl.com/xunlian/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
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> 在哈希法中有一些场景就是为数组量身定做的。
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# 383. 赎金信
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[力扣题目链接](https://leetcode.cn/problems/ransom-note/)
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给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串 ransom 能不能由第二个字符串 magazines 里面的字符构成。如果可以构成，返回 true ；否则返回 false。
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(题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思。杂志字符串中的每个字符只能在赎金信字符串中使用一次。)
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**注意：**
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你可以假设两个字符串均只含有小写字母。
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canConstruct("a", "b") -> false
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canConstruct("aa", "ab") -> false
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canConstruct("aa", "aab") -> true
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## 思路
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这道题目和[242.有效的字母异位词](https://programmercarl.com/0242.有效的字母异位词.html)很像，[242.有效的字母异位词](https://programmercarl.com/0242.有效的字母异位词.html)相当于求 字符串a 和 字符串b 是否可以相互组成 ，而这道题目是求 字符串a能否组成字符串b，而不用管字符串b 能不能组成字符串a。
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本题判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成，但是这里需要注意两点。
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* 第一点“为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思” 这里*说明杂志里面的字母不可重复使用。*
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* 第二点 “你可以假设两个字符串均只含有小写字母。” *说明只有小写字母*，这一点很重要
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### 暴力解法
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那么第一个思路其实就是暴力枚举了，两层for循环，不断去寻找，代码如下：
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```CPP
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class Solution {
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public:
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bool canConstruct(string ransomNote, string magazine) {
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for (int i = 0; i < magazine.length(); i++) {
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for (int j = 0; j < ransomNote.length(); j++) {
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// 在ransomNote中找到和magazine相同的字符
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if (magazine[i] == ransomNote[j]) {
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ransomNote.erase(ransomNote.begin() + j); // ransomNote删除这个字符
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break;
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}
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}
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}
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// 如果ransomNote为空，则说明magazine的字符可以组成ransomNote
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if (ransomNote.length() == 0) {
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return true;
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}
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return false;
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}
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};
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```
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* 时间复杂度: O(n^2)
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* 空间复杂度: O(1)
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这里时间复杂度是比较高的，而且里面还有一个字符串删除也就是erase的操作，也是费时的，当然这段代码也可以过这道题。
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### 哈希解法
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因为题目说只有小写字母，那可以采用空间换取时间的哈希策略，用一个长度为26的数组来记录magazine里字母出现的次数。
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然后再用ransomNote去验证这个数组是否包含了ransomNote所需要的所有字母。
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依然是数组在哈希法中的应用。
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一些同学可能想，用数组干啥，都用map完事了，**其实在本题的情况下，使用map的空间消耗要比数组大一些的，因为map要维护红黑树或者哈希表，而且还要做哈希函数，是费时的！数据量大的话就能体现出来差别了。 所以数组更加简单直接有效！**
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代码如下：
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```CPP
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class Solution {
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public:
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bool canConstruct(string ransomNote, string magazine) {
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int record[26] = {0};
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//add
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if (ransomNote.size() > magazine.size()) {
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return false;
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}
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for (int i = 0; i < magazine.length(); i++) {
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// 通过record数据记录 magazine里各个字符出现次数
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record[magazine[i]-'a'] ++;
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}
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for (int j = 0; j < ransomNote.length(); j++) {
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// 遍历ransomNote，在record里对应的字符个数做--操作
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record[ransomNote[j]-'a']--;
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// 如果小于零说明ransomNote里出现的字符，magazine没有
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if(record[ransomNote[j]-'a'] < 0) {
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return false;
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}
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}
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return true;
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}
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};
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```
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* 时间复杂度: O(n)
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* 空间复杂度: O(1)
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## 其他语言版本
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### Java：
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```Java
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class Solution {
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public boolean canConstruct(String ransomNote, String magazine) {
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// shortcut
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if (ransomNote.length() > magazine.length()) {
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return false;
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}
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// 定义一个哈希映射数组
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int[] record = new int[26];
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// 遍历
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for(char c : magazine.toCharArray()){
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record[c - 'a'] += 1;
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}
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for(char c : ransomNote.toCharArray()){
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record[c - 'a'] -= 1;
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}
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// 如果数组中存在负数，说明ransomNote字符串中存在magazine中没有的字符
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for(int i : record){
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if(i < 0){
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return false;
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}
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}
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return true;
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}
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}
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```
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### Python：
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（版本一）使用数组
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```python
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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ransom_count = [0] * 26
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magazine_count = [0] * 26
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for c in ransomNote:
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ransom_count[ord(c) - ord('a')] += 1
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for c in magazine:
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magazine_count[ord(c) - ord('a')] += 1
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return all(ransom_count[i] <= magazine_count[i] for i in range(26))
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```
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（版本二）使用defaultdict
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```python
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from collections import defaultdict
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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hashmap = defaultdict(int)
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for x in magazine:
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hashmap[x] += 1
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for x in ransomNote:
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value = hashmap.get(x)
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if not value or not value:
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return False
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else:
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hashmap[x] -= 1
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return True
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```
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（版本三）使用字典
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```python
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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counts = {}
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for c in magazine:
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counts[c] = counts.get(c, 0) + 1
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for c in ransomNote:
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if c not in counts or counts[c] == 0:
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return False
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counts[c] -= 1
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return True
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```
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（版本四）使用Counter
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```python
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from collections import Counter
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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return not Counter(ransomNote) - Counter(magazine)
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```
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（版本五）使用count
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```python
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote))
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```
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(版本六）使用count(简单易懂)
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```python
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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for char in ransomNote:
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if char in magazine and ransomNote.count(char) <= magazine.count(char):
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continue
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else:
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return False
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return True
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```
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### Go：
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```go
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func canConstruct(ransomNote string, magazine string) bool {
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record := make([]int, 26)
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for _, v := range magazine { // 通过record数据记录 magazine里各个字符出现次数
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record[v-'a']++
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}
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for _, v := range ransomNote { // 遍历ransomNote，在record里对应的字符个数做--操作
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record[v-'a']--
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if record[v-'a'] < 0 { // 如果小于零说明ransomNote里出现的字符，magazine没有
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return false
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}
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}
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return true
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}
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```
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### JavaScript:
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```js
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/**
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* @param {string} ransomNote
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* @param {string} magazine
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* @return {boolean}
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*/
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var canConstruct = function(ransomNote, magazine) {
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const strArr = new Array(26).fill(0),
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base = "a".charCodeAt();
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for(const s of magazine) { // 记录 magazine里各个字符出现次数
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strArr[s.charCodeAt() - base]++;
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}
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for(const s of ransomNote) { // 对应的字符个数做--操作
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const index = s.charCodeAt() - base;
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if(!strArr[index]) return false; // 如果没记录过直接返回false
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strArr[index]--;
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}
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return true;
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};
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```
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### TypeScript:
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```typescript
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function canConstruct(ransomNote: string, magazine: string): boolean {
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let helperArr: number[] = new Array(26).fill(0);
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let base: number = 'a'.charCodeAt(0);
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let index: number;
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for (let i = 0, length = magazine.length; i < length; i++) {
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helperArr[magazine[i].charCodeAt(0) - base]++;
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}
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for (let i = 0, length = ransomNote.length; i < length; i++) {
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index = ransomNote[i].charCodeAt(0) - base;
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helperArr[index]--;
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if (helperArr[index] < 0) {
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return false;
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}
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}
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return true;
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};
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```
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### PHP:
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```php
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class Solution {
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/**
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* @param String $ransomNote
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* @param String $magazine
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* @return Boolean
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*/
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function canConstruct($ransomNote, $magazine) {
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if (count($ransomNote) > count($magazine)) {
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return false;
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}
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$map = [];
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for ($i = 0; $i < strlen($magazine); $i++) {
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$map[$magazine[$i]] = ($map[$magazine[$i]] ?? 0) + 1;
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}
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for ($i = 0; $i < strlen($ransomNote); $i++) {
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if (!isset($map[$ransomNote[$i]]) || --$map[$ransomNote[$i]] < 0) {
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return false;
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}
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}
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return true;
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}
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```
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### Swift：
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```swift
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func canConstruct(_ ransomNote: String, _ magazine: String) -> Bool {
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var record = Array(repeating: 0, count: 26);
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let aUnicodeScalarValue = "a".unicodeScalars.first!.value
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for unicodeScalar in magazine.unicodeScalars {
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// 通过record 记录 magazine 里各个字符出现的次数
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let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
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record[idx] += 1
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}
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for unicodeScalar in ransomNote.unicodeScalars {
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// 遍历 ransomNote,在record里对应的字符个数做 -- 操作
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let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
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record[idx] -= 1
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// 如果小于零说明在magazine没有
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if record[idx] < 0 {
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return false
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}
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}
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return true
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}
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```
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### Rust:
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```rust
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impl Solution {
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pub fn can_construct(ransom_note: String, magazine: String) -> bool {
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let baseChar = 'a';
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let mut record = vec![0; 26];
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for byte in magazine.bytes() {
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record[byte as usize - baseChar as usize] += 1;
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}
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for byte in ransom_note.bytes() {
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record[byte as usize - baseChar as usize] -= 1;
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if record[byte as usize - baseChar as usize] < 0 {
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return false;
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}
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}
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return true;
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}
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}
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```
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### Scala:
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版本一: 使用数组作为哈希表
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```scala
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object Solution {
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def canConstruct(ransomNote: String, magazine: String): Boolean = {
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// 如果magazine的长度小于ransomNote的长度，必然是false
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if (magazine.length < ransomNote.length) {
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return false
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}
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// 定义一个数组，存储magazine字符出现的次数
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val map: Array[Int] = new Array[Int](26)
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// 遍历magazine字符串，对应的字符+=1
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for (i <- magazine.indices) {
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map(magazine(i) - 'a') += 1
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}
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// 遍历ransomNote
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for (i <- ransomNote.indices) {
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if (map(ransomNote(i) - 'a') > 0)
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map(ransomNote(i) - 'a') -= 1
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else return false
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}
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// 如果上面没有返回false，直接返回true，关键字return可以省略
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true
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}
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}
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```
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```scala
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object Solution {
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import scala.collection.mutable
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def canConstruct(ransomNote: String, magazine: String): Boolean = {
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// 如果magazine的长度小于ransomNote的长度，必然是false
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if (magazine.length < ransomNote.length) {
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return false
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}
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// 定义map，key是字符，value是字符出现的次数
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val map = new mutable.HashMap[Char, Int]()
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// 遍历magazine，把所有的字符都记录到map里面
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for (i <- magazine.indices) {
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val tmpChar = magazine(i)
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// 如果map包含该字符，那么对应的value++，否则添加该字符
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if (map.contains(tmpChar)) {
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map.put(tmpChar, map.get(tmpChar).get + 1)
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} else {
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map.put(tmpChar, 1)
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}
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}
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// 遍历ransomNote
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for (i <- ransomNote.indices) {
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val tmpChar = ransomNote(i)
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// 如果map包含并且该字符的value大于0，则匹配成功，map对应的--，否则直接返回false
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if (map.contains(tmpChar) && map.get(tmpChar).get > 0) {
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map.put(tmpChar, map.get(tmpChar).get - 1)
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} else {
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return false
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}
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}
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// 如果上面没有返回false，直接返回true，关键字return可以省略
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true
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}
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}
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```
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### C#：
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```csharp
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public bool CanConstruct(string ransomNote, string magazine) {
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if(ransomNote.Length > magazine.Length) return false;
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int[] letters = new int[26];
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foreach(char c in magazine){
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letters[c-'a']++;
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}
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foreach(char c in ransomNote){
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letters[c-'a']--;
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if(letters[c-'a']<0){
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return false;
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}
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}
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return true;
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}
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```
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### C：
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```c
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bool canConstruct(char* ransomNote, char* magazine) {
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// 定义哈希映射数组
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int hashmap[26] = {0};
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// 对magazine中字符计数
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while (*magazine != '\0') hashmap[*magazine++ % 26]++;
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// 遍历ransomNote，对应的字符自减，小于0说明该字符magazine没有或不足够表示
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while (*ransomNote != '\0') hashmap[*ransomNote++ % 26]--;
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// 如果数组中存在负数，说明ransomNote不能由magazine里面的字符构成
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for (int i = 0; i < 26; i++) {
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if (hashmap[i] < 0) return false;
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}
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return true;
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}
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```
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<p align="center">
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||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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