leetcode322. Coin Change

最新推荐文章于 2024-06-06 11:29:05 发布

原创最新推荐文章于 2024-06-06 11:29:05 发布 · 193 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#dp #背包

—dp 同时被 3 个专栏收录

152 篇文章

订阅专栏

leetcode

89 篇文章

订阅专栏

———背包

33 篇文章

订阅专栏

本文深入探讨了完全背包问题的解决方案，通过LeetCode上的经典题目“Coin Change”进行讲解。使用C++实现动态规划算法，计算出构成指定金额所需的最少硬币数量。若无法构成，则返回-1。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

辣鸡如我，居然现在完全背包还要想这么久

https://leetcode.com/problems/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:

Input: coins = [1, 2, 5], amount = 11
Output: 3 
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int dp[20000];
        for (int i = 1; i < 20000; i ++)
            dp[i] = 0x3f3f3f3f;
        dp[0] = 0;
        for (int i = 1; i <= amount ; i ++) {
            for (int j = 0; j < coins.size() ; j ++) {
                if (coins[j]  <= i) {
                    dp[i] = min(dp[i - coins[j]] + 1, dp[i]);
                }
            }
        }
        if (dp[amount] != 0x3f3f3f3f)
            return dp[amount];
        else
            return -1;
    }
};