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:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)
Example 2:
coins = [2], amount = 3
return -1.
Note:
You may assume that you have an infinite number of each kind of coin.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
其中ans[i]表示钱数为i时的最小硬币数的找零，递推式为：
ans[i] = Math.min(ans[i], ans[i - coins[j]] + 1);
其中coins[j]为第j个硬币，而i - coins[j]为钱数i减去其中一个硬币的值，剩余的钱数在ans数组中找到值，然后加1和当前ans数组中的值做比较，取较小的那个更新ans数组。
1 public class Solution { 2 public int coinChange(int[] coins, int amount) { 3 int[] ans = new int[amount+1]; 4 Arrays.fill(ans,amount+1); 5 ans[0] = 0; 6 for(int i = 1; i < ans.length; i++){ 7 for(int j = 0; j < coins.length; j++){ 8 if(coins[j] <= i) 9 ans[i] = Math.min(ans[i],ans[i-coins[j]]+1); 10 } 11 } 12 return ans[amount] == (amount+1) ? -1 : ans[amount]; 13 } 14 }