leetcode-322硬币找零

最新推荐文章于 2021-12-14 11:37:37 发布

xiepengcheng716

最新推荐文章于 2021-12-14 11:37:37 发布

阅读量625

点赞数

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/u010732356/article/details/70807634

版权

leetcode 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种解决找零问题的高效算法，旨在计算组成特定金额所需的最少硬币数量。通过使用动态规划方法，该算法能在多项式时间内找到最优解，并处理无法找零的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

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

        int Max = amount + 1;
        vector<int> dp(amount + 1, Max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < coins.size(); j++) {322
                if (coins[j] <= i) {
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};