322--Coin Change比较清晰的动态规划,状态转移方程和起始状态都是比较好找到的,但需要一系列的学习才能对这类动态规划问题熟悉。我会继续上传这个问题的变式的解法。, 视频播放量 49、弹幕量 0、点赞数 2、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 Nemesiscs, 作
【Leetcode】Coin Change 题目链接:https://leetcode.com/problems/coin-change/题目: -1.Example 1: coins =[1, 2, 5], amount =11 return3Example 2: coins =[2], amount =3 return-1. Note: You may assume that you have an infinite number of each kind of coin. 思路: c[i]表示数目为i...
https://leetcode-cn.com/problems/coin-change 解题思路 动态规划,自底向上,太简单,不解释。 C++代码 classSolution{public:intcoinChange(vector<int>& coins,intamount){vector<int>dp(amount +1, amount +1); dp[0] =0;for(inti =1; i <= amount; i++) {for(autov : coins) {if(i >= v)...
Returnthe 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. You may assume that you have an infinite number of each kind of coin. Example 1: Input:coins = [1,2,5], amount = 11Output...
题目地址:https://leetcode.com/problems/coin-change/description/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 ...
https://leetcode.cn/problems/coin-changeleetcode.cn/problems/coin-change dp解法: 定义dp矩阵:dp[j]为凑成总金额为j时最少的硬币个数,则dp[amount]为我们最终要求的结果 定义dp转移方程:假设coins有coin_1、coin_2、...、coin_n,则dp[i] = min{dp[i-coin_1], dp[i-coin_2], ... ,dp[...
示例2: 输入:coins = [2], amount = 3 输出:-1 示例3: 输入:coins = [1], amount = 0 输出:0 提示: 1 <= coins.length <= 12 1 <= coins[i] <= 231 - 1 0 <= amount <= 104 题目难度:中等 通过次数:956.8K 提交次数:1.9M 贡献者:LeetCode 相关标签 相似题目 C++...
题目地址:https://leetcode.com/problems/coin-change/description/ 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 ...
def coinChange(self, coins: [int], amount: int) -> int: # 声明一个二维矩阵 matrix = [[i for i in range(amount + 1)] for _ in range(2)] # 初始化第一行 for i in range(amount + 1): # 如果当前位置表示的需要兑换的钱数可以被整除,将当前位置置为需要钱的个数 ...
1 要注意corner case:当amount==0的时候,返回0,表示有0种组合方式 2 还有就是如果amount在coins中,直接返回1 3 queue中压栈money和dist 4 重点在什么时候加到visited中去,只有没有在visited中,才加入到visited中和放入队列中去 1当amount是0的时候,返回0,不是-1 ...