第十七周 leetcode 322. Coin Change（Medium）

题目描述：

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.

解题思路：

动态规划的思路：
dp[i]表示组成数字i需要的最少数量。
状态转移方程：dp[i]=min(dp[i-coins[j]]+1,dp[i]);

代码：

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

代码运行结果：