LeetCode 322. 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.

分析：动态规划。dp[i]表示钱数为i时找零的最小的个数。
$dp[i] = min\{dp[i], dp[i- coins[j]] + 1\}, 0 \leq k < n$
边界条件：
1）amount=0
2）coins为空
代码如下：

class Solution {
public:
    int coinChange(vector<int> &coins,int amount)
    {
        if(amount==0) return 0;
        if(coins.size()==0) return -1;
        const int maxn = amount+1;

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

代码2：大致思路一样，实现不一样

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int result=count(amount,coins);
        return result==INT_MAX? -1 : result;
    }
    int count(int n,vector<int> &coins) {
        if(n==0) return 0;
        if(coins.size()==0) return -1;

        int dp[n+1]={0};
        for(int i=1;i<=n;i++) {
            int minVal=INT_MAX;
            for(int j=0;j<coins.size();j++) {
                if(i>=coins[j] && dp[i-coins[j]]!=-1) {
                    minVal=min(minVal,dp[i-coins[j]]+1);
                }
            }
            dp[i]=minVal==INT_MAX?-1:minVal;
        }

        return dp[n];
    }
};