参与本项目，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收益！

partitionLabels(string S) { int hash[27] = {0}; // i为字符，hash[i]为字符出现的最后位置 for (int i = 0; i < S.size(); i++) { // 统计每一个字符最后出现的位置 hash[S[i] - 'a'] = i; } vector result; int left = 0; int right = 0; for (int i = 0; i < S.size(); i++) { right = max(right, hash[S[i] - 'a']); // 找到字符出现的最远边界 if (i == right) { result.push_back(right - left + 1); left = i + 1; } } return result; } }; ``` * 时间复杂度：$O(n)$ * 空间复杂度：$O(1)$，使用的hash数组是固定大小 ## 总结 这道题目leetcode标记为贪心算法，说实话，我没有感受到贪心，找不出局部最优推出全局最优的过程。就是用最远出现距离模拟了圈字符的行为。 但这道题目的思路是很巧妙的，所以有必要介绍给大家做一做，感受一下。 ## 补充 这里提供一种与[452.用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)、[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)相同的思路。 统计字符串中所有字符的起始和结束位置，记录这些区间(实际上也就是[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)题目里的输入)，**将区间按左边界从小到大排序，找到边界将区间划分成组，互不重叠。找到的边界就是答案。** ```CPP class Solution { public: static bool cmp(vector &a, vector &b) { return a[0] < b[0]; } // 记录每个字母出现的区间 vector> countLabels(string s) { vector> hash(26, vector(2, INT_MIN)); vector> hash_filter; for (int i = 0; i < s.size(); ++i) { if (hash[s[i] - 'a'][0] == INT_MIN) { hash[s[i] - 'a'][0] = i; } hash[s[i] - 'a'][1] = i; } // 去除字符串中未出现的字母所占用区间 for (int i = 0; i < hash.size(); ++i) { if (hash[i][0] != INT_MIN) { hash_filter.push_back(hash[i]); } } return hash_filter; } vector partitionLabels(string s) { vector res; // 这一步得到的 hash 即为无重叠区间题意中的输入样例格式：区间列表 // 只不过现在我们要求的是区间分割点 vector> hash = countLabels(s); // 按照左边界从小到大排序 sort(hash.begin(), hash.end(), cmp); // 记录最大右边界 int rightBoard = hash[0][1]; int leftBoard = 0; for (int i = 1; i < hash.size(); ++i) { // 由于字符串一定能分割，因此, // 一旦下一区间左边界大于当前右边界，即可认为出现分割点 if (hash[i][0] > rightBoard) { res.push_back(rightBoard - leftBoard + 1); leftBoard = hash[i][0]; } rightBoard = max(rightBoard, hash[i][1]); } // 最右端 res.push_back(rightBoard - leftBoard + 1); return res; } }; ``` ## 其他语言版本 ### Java ```java class Solution { public List partitionLabels(String S) { List list = new LinkedList<>(); int[] edge = new int[26]; char[] chars = S.toCharArray(); for (int i = 0; i < chars.length; i++) { edge[chars[i] - 'a'] = i; } int idx = 0; int last = -1; for (int i = 0; i < chars.length; i++) { idx = Math.max(idx,edge[chars[i] - 'a']); if (i == idx) { list.add(i - last); last = i; } } return list; } } class Solution{ /*解法二： 上述c++补充思路的Java代码实现*/ public int[][] findPartitions(String s) { List temp = new ArrayList<>(); int[][] hash = new int[26][2];//26个字母2列 表示该字母对应的区间 for (int i = 0; i < s.length(); i++) { //更新字符c对应的位置i char c = s.charAt(i); if (hash[c - 'a'][0] == 0) hash[c - 'a'][0] = i; hash[c - 'a'][1] = i; //第一个元素区别对待一下 hash[s.charAt(0) - 'a'][0] = 0; } List> h = new LinkedList<>(); //组装区间 for (int i = 0; i < 26; i++) { //if (hash[i][0] != hash[i][1]) { temp.clear(); temp.add(hash[i][0]); temp.add(hash[i][1]); //System.out.println(temp); h.add(new ArrayList<>(temp)); // } } // System.out.println(h); // System.out.println(h.size()); int[][] res = new int[h.size()][2]; for (int i = 0; i < h.size(); i++) { List list = h.get(i); res[i][0] = list.get(0); res[i][1] = list.get(1); } return res; } public List partitionLabels(String s) { int[][] partitions = findPartitions(s); List res = new ArrayList<>(); Arrays.sort(partitions, (o1, o2) -> Integer.compare(o1[0], o2[0])); int right = partitions[0][1]; int left = 0; for (int i = 0; i < partitions.length; i++) { if (partitions[i][0] > right) { //左边界大于右边界即可纪委一次分割 res.add(right - left + 1); left = partitions[i][0]; } right = Math.max(right, partitions[i][1]); } //最右端 res.add(right - left + 1); return res; } } ``` ### Python ```python class Solution: def partitionLabels(self, s: str) -> List[int]: hash = [0] * 26 for i in range(len(s)): hash[ord(s[i]) - ord('a')] = i result = [] left = 0 right = 0 for i in range(len(s)): right = max(right, hash[ord(s[i]) - ord('a')]) if i == right: result.append(right - left + 1) left = i + 1 return result ``` ### Go ```go func partitionLabels(s string) []int { var res []int; var marks [26]int; size, left, right := len(s), 0, 0; for i := 0; i < size; i++ { marks[s[i] - 'a'] = i; } for i := 0; i < size; i++ { right = max(right, marks[s[i] - 'a']); if i == right { res = append(res, right - left + 1); left = i + 1; } } return res; } func max(a, b int) int { if a < b { a = b; } return a; } ``` ### Javascript ```Javascript var partitionLabels = function(s) { let hash = {} for(let i = 0; i < s.length; i++) { hash[s[i]] = i } let result = [] let left = 0 let right = 0 for(let i = 0; i < s.length; i++) { right = Math.max(right, hash[s[i]]) if(right === i) { result.push(right - left + 1) left = i + 1 } } return result }; ``` ### TypeScript ```typescript function partitionLabels(s: string): number[] { const length: number = s.length; const resArr: number[] = []; const helperMap: Map = new Map(); for (let i = 0; i < length; i++) { helperMap.set(s[i], i); } let left: number = 0; let right: number = 0; for (let i = 0; i < length; i++) { right = Math.max(helperMap.get(s[i])!, right); if (i === right) { resArr.push(i - left + 1); left = i + 1; } } return resArr; }; ``` ### Scala ```scala object Solution { import scala.collection.mutable def partitionLabels(s: String): List[Int] = { var hash = new Array[Int](26) for (i <- s.indices) { hash(s(i) - 'a') = i } var res = mutable.ListBuffer[Int]() var (left, right) = (0, 0) for (i <- s.indices) { right = math.max(hash(s(i) - 'a'), right) if (i == right) { res.append(right - left + 1) left = i + 1 } } res.toList } } ``` ### Rust ```Rust use std::collections::HashMap; impl Solution { fn max (a: usize, b: usize) -> usize { if a > b { a } else { b } } pub fn partition_labels(s: String) -> Vec { let s = s.chars().collect::>(); let mut hash: HashMap = HashMap::new(); for i in 0..s.len() { hash.insert(s[i], i); } let mut result: Vec = Vec::new(); let mut left: usize = 0; let mut right: usize = 0; for i in 0..s.len() { right = Self::max(right, hash[&s[i]]); if i == right { result.push((right - left + 1) as i32); left = i + 1; } } result } } ``` -----------------------