LeetCode Coin Change

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.

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        vector<int> dp(amount+1, amount+1);
        dp[0] = 0;
        for(int i = 0; i<n; i++){
            for(int j = 1; j<=amount; j++){
                if(j >= coins[i])
                    dp[j] = min(1 + dp[j-coins[i]], dp[j]);
            }
        }
        return dp[amount] > amount ? -1: dp[amount];
    }
};

巧妙之处：
把数组初始化为amount+1， 这个数可以看成无穷大， 因为只要能找零， 结果一定比amount+1小。

滚动数组避免了不必要的赋值。