322. Coin Change 动态规划应用

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.

分析：

典型的dp问题，以count=5,coins={1,2,5}分析。dp[i]表示第i个货币最小需要dp[i]个货币。我们从0开始，0需要0个硬币，1需要一个1元货币。dp[1]=1。2有两个方案，一个1元再加另一个一元。dp[2]=dp[1]+1=2，另一种方案是一个2元,所以dp[2]=1。dp[3]也有两种方案dp[1]+dp[2]或dp[1]+dp[1]+dp[1]。所以dp[3]=2。以此类推，在我的代码中，初始化是用INT_MAX。要注意溢出。

代码：

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