LeetCode #322 - Coin Change_you are given coins of di erent denominations and -优快云博客

本文介绍了一种使用动态规划解决硬币找零问题的方法。针对给定的不同面额的硬币集合和目标金额，该算法能够计算出构成该金额所需的最少硬币数量。若无法通过现有硬币组合得到目标金额，则返回-1。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

给出硬币面值和某一金额，求最少需要多少个硬币才能凑足，硬币数量不限。

采用动态规划，对于i，如果i>=coins[j]&&dp[i-coins[j]]>0，那么金额i可以按硬币面值兑换，遍历满足条件的coins[j]，找出最少的硬币数即可。

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