Coin Change

题目来源
一道简单的DP题，n种硬币，要求组成某个数值的硬币数最少，代码如下：

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

看了下讨论区，感觉可以写的更简洁一下，代码如下：

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