leetcode之 coin change

最新推荐文章于 2025-06-03 13:27:00 发布

luckyu1

最新推荐文章于 2025-06-03 13:27:00 发布

阅读量304

点赞数

CC 4.0 BY-SA版权

文章标签： leetcode 算法面试

本文链接：https://blog.youkuaiyun.com/luckyu1/article/details/50981233

本文介绍了一个典型的动态规划问题——硬币找零问题，并提供了一种使用C++实现的有效解决方案。该问题旨在寻找构成特定金额所需的最少硬币数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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.

解答：

很典型的Dp，直接代码即可

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> record;
        record.push_back(0);
        sort(coins.begin(),coins.end());
        for(int i = 1; i <= amount; ++i)
        {
            int tmp = 1 << 30;
            for(int j = 0;j < coins.size() && coins[j] <= i;++j)
            {
                if(1 + record[i - coins[j]] < tmp)
                    tmp = 1 + record[i - coins[j]];
            }
            record.push_back(tmp);
        }
        return record[amount] == 1 << 30 ? -1 : record[amount];
    }
};