79 lines
2.5 KiB
Markdown
79 lines
2.5 KiB
Markdown
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[1] 0 ，输出的是0，不是-1啊，这颗真是天坑j
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```
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// dp初始化很重要
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class Solution {
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public:
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int coinChange(vector<int>& coins, int amount) {
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//int dp[10003] = {0}; // 并没有给所有元素赋值0
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if (amount == 0) return 0; // 这个要注意
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vector<int> dp(10003, 0);
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// 不能这么初始化啊，[2147483647]，2 这种例子 直接gg，但是这种初始化有助于理解
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for (int i = 0; i < coins.size(); i++) {
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if (coins[i] <= amount) // 还必须要加这个判断
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dp[coins[i]] = 1;
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}
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for (int i = 1; i <= amount; i++) {
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for (int j = 0; j < coins.size(); j++) {
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if (i - coins[j] >= 0 && dp[i - coins[j]]!=0 ) {
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if (dp[i] == 0) dp[i] = dp[i - coins[j]] + 1;
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else dp[i] = min(dp[i - coins[j]] + 1, dp[i]);
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}
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}
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//for (int k = 0 ; k<= amount; k++) {
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// cout << dp[k] << " ";
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//}
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//cout << endl;
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}
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if (dp[amount] == 0) return -1;
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return dp[amount];
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}
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};
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```
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我用求组合的思路也过了，
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```
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class Solution {
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public:
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int coinChange(vector<int>& coins, int amount) {
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//int dp[10003] = {0}; // 并没有给所有元素赋值0
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//if (amount == 0) return 0;
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vector<int> dp(10003, INT_MAX);
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dp[0] = 0;
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for (int i = 0 ;i < coins.size(); i++) { // 求组合
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for (int j = 1; j <= amount; j++) {
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if (j - coins[i] >= 0 && dp[j - coins[i]] != INT_MAX) {
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dp[j] = min(dp[j - coins[i]] + 1, dp[j]);
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}
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}
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}
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if (dp[amount] == INT_MAX) return -1;
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return dp[amount];
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}
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};
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```
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这种标记d代码简短，但思路有点绕
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```
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class Solution {
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public:
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int coinChange(vector<int>& coins, int amount) {
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//int dp[10003] = {0}; // 并没有给所有元素赋值0
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// if (amount == 0) return 0; 这个都可以省略了，但很多同学不知道 还需要注意这个
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vector<int> dp(10003, 0);
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for (int i = 1; i <= amount; i++) {
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dp[i] = INT_MAX;
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for (int j = 0; j < coins.size(); j++) {
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if (i - coins[j] >= 0 && dp[i - coins[j]]!=INT_MAX ) {
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dp[i] = min(dp[i - coins[j]] + 1, dp[i]);
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}
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}
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}
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if (dp[amount] == INT_MAX) return -1;
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return dp[amount];
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}
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};
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```
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