Leetcode322. Coin Change

最新推荐文章于 2024-06-06 11:29:05 发布

没事多喝白开水

最新推荐文章于 2024-06-06 11:29:05 发布

阅读量119

点赞数

CC 4.0 BY-SA版权

分类专栏： C++ Leetcode

本文链接：https://blog.youkuaiyun.com/kuaijg/article/details/98512559

C++ 同时被 2 个专栏收录

69 篇文章

订阅专栏

Leetcode

49 篇文章

订阅专栏

本文探讨了硬币找零问题，即如何使用最少数量的硬币组合来构成特定金额。通过动态规划的方法，我们提供了一个44ms的解决方案，详细解释了算法的实现过程和核心思路。

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:

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

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

动态规划，44ms

class Solution {
public:
	int coinChange(vector<int>& coins, int amount) {
		int len = coins.size();
		if (amount == 0) {
			return 0;
		}
		if (len == 0 ) {
			return -1;
		}
		vector<int> dp(amount+1, INT_MAX);
		dp[0] = 0;
		for (int i = 1; i <= amount; ++i) {
			for (int coin : coins) {
				if (coin <= amount) {
					dp[coin] = 1;
				}
				if (i- coin >=0&&dp[i - coin] < dp[i]-1 ) {
					dp[i] = dp[i- coin] + 1;
				}
			}
		}
		return dp[amount] == INT_MAX ? -1: dp[amount];
	}
};