贵州建站管理系统网络营销方式优势

题目链接：322. 零钱兑换 - 力扣（LeetCode）

这道题目使用动态规划

状态转移方程：

dp[j] = min(dp[j], dp[j-coins[i]]+1);

但是做这道题目遇到些小问题，这里提一下

1、一开始是将dp初始化为INT_MAX的。但提交会报错

int min(int a, int b){return a >= b ? b : a;
}int coinChange(int* coins, int coinsSize, int amount) {int dp[amount+1];for (int i = 0; i <= amount; i++) {dp[i] = INT_MAX;}dp[0] = 0;for (int i = 0; i < coinsSize; i++) {for (int j = coins[i]; j <= amount; j++) {if (j-coins[i] >= 0) {dp[j] = min(dp[j], dp[j-coins[i]]+1);}}}if (dp[amount] == INT_MAX)return -1;elsereturn dp[amount];
}

报错截图

遂将dp初始化为amount，但只通过179 / 189，原因是amount为0时，答案错误

 

最后将dp初始化为amount+1，通过

int min(int a, int b){return a >= b ? b : a;
}int coinChange(int* coins, int coinsSize, int amount) {int dp[amount+1];for (int i = 0; i <= amount; i++) {dp[i] = amount+1;}dp[0] = 0;for (int i = 0; i < coinsSize; i++) {for (int j = coins[i]; j <= amount; j++) {if (j-coins[i] >= 0) {dp[j] = min(dp[j], dp[j-coins[i]]+1);}}}if (dp[amount] == amount+1)return -1;elsereturn dp[amount];
}