322.CoinChange

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Note:

You may assume that you have an infinite number of each kind of coin.

ans[i]表示凑齐钱数i时需要的最小零钱数目

和题目494类似，但是内层遍历要从小到大

public int coinChange(int[] coins, int amount) {
  int[] ans = new int[amount + 1];
  for (int i = 0; i < amount + 1; i++)
    ans[i] = amount + 1;//注意，初始化的时候不能设成Integer.MAX_VALUE,后面要有ans[i-coins[j]]+1的操作，会溢出称为Integer.MIN_VALUE
  ans[0] = 0;
  for (int i = 0; i < coins.length; i++) {
    for (int j = coins[i]; j <= amount; j++) {
        ans[j] = Math.min(ans[j], ans[j - coins[i]] + 1);
    }
  }
  return ans[amount] == (amount + 1) ? -1 : ans[amount];
}