LeetCode 322. Coin Change

322. Coin Change

Description
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.

Analysis
这道题的意思就是给定一个向量表示当前零钱，再给定一个整数当做所给定的钱。
要求求零钱凑成这个整数所需要的最少数量，若没有办法凑成，则返回-1。
我的做法是利用动态规划。
其实凑成当前钱数所需要的最少数量，就等于当前钱数减去某一个零钱数得到的整数所需要的最少数量+1。
注意有可能没有办法凑成当前钱数，当其减去当前钱数减去某一个零钱数得到的整数是没有办法凑成的。意味着它利用当前的钱数凑成方法也没有办法构成。
进行循环就可以得到结构。

Code

class Solution {public:    int coinChange(vector<int>& coins, int amount) {        int dp[amount+1];        for(int i=1;i<=amount;++i){            dp[i]=-1;        }        dp[0]=0;        for(int i=0;i<coins.size();++i){            if(coins[i]<=amount)            dp[coins[i]]=1;        }        for(int i=1;i<=amount;++i){            int res=100000;            for(int j =0;j<coins.size();++j){                if(i-coins[j]>=0&&dp[i-coins[j]] == -1) dp[i]=-1;                else if(i-coins[j]>=0) res=min(res,dp[i-coins[j]]+1);            }            if(res!=100000) dp[i]=res;        }        return dp[amount];    }};