标签：|| style mon tco put not cti NPU one
You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
Example 1:
Input: coins =[1, 2, 5], amount =11Output:3Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins =[2], amount =3Output: -1
Note:
You may assume that you have an infinite number of each kind of coin.
使用dp，
1 public int coinChange(int[] coins, int amount) {//dp my 2 if(null==coins||0==coins.length){ 3 return -1; 4 } 5 int[] re = new int[amount+1];//re[i]表示总额为i需要至少多上个零钱 6 for(int i=1;i<=amount;i++){ 7 re[i] = Integer.MAX_VALUE; 8 for(int j =0;j<coins.length;j++){//re[i]=min(re[i-coins[j]]+1) 9 if(coins[j]<=i&&re[i]>re[i-coins[j]]){ 10 re[i] = re[i-coins[j]]; 11 } 12 } 13 if(re[i]!=Integer.MAX_VALUE){ 14 re[i]++; 15 } 16 17 } 18 if(re[amount]==Integer.MAX_VALUE){ 19 return -1; 20 } 21 return re[amount]; 22 }
标签：|| style mon tco put not cti NPU one
原文地址：https://www.cnblogs.com/zhacai/p/10661489.html
