依次向前推,直到r等于0或者小于0. public int coinChange(int[] coins, int amount) { if (amount < 0) return 0; return coinChangeCore(coins, amount, new int[amount]); } private int coinChangeCore(int[] coins, int amount, int[] count) { if (amount < 0) return -1; if (amount ==...
Java实现 1classSolution {2publicintcoinChange(int[] coins,intamount) {3//memo[n]的值: 表示的凑成总金额为n所需的最少的硬币个数4int[] memo =newint[amount + 1];5memo[0] = 0;6for(inti = 1; i <= amount; i++) {7intmin =Integer.MAX_VALUE;8for(intj = 0; j < coins.length...
322. Coin ChangeMedium Topics Companies You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made...
图片来自于https://leetcode-cn.com/problems/coin-change/solution/wan-quan-bei-bao-wen-ti-shou-hua-dp-table-by-shixu/ 值得一提的是, 以原问题 amount = 11, coins = [2,5,1] 为例子。 i=1代表的是只可以放 {coin = 2,count =1}下,可以放n个小于11的情况 ...
leetcode的testcase有一个为amount为0的时候,所以在代码的开头加了一个判断。 吐槽一下leetcode怎么有这么多诡异的testcase。 classSolution:defcoinChange(self,coins:List[int],amount:int)->int:ifamount==0:return0dp=[amount+1]*(amount+1)dp[0]=0foriinrange(1,amount+1):forcincoins:ifi>=c:dp...
LeetCode 322. Coin Change 简介:给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额,返回 -1。 Description \You are given coins of different denominations and a total amount of money amount. Write a ...
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
package leetcode func coinChange(coins []int, amount int) int { dp := make([]int, amount+1) dp[0] = 0 for i := 1; i < len(dp); i++ { dp[i] = amount + 1 } for i := 1; i <= amount; i++ { for j := 0; j < len(coins); j++ { ...
leetcode322. Coin Change 题目要求 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, ...
LeetCode 322. Coin Change 题目 动态规划 class Solution { public: int dp[10005]; int coinChange(vector<int>& coins, int amount) { memset(dp,-1,sizeof(dp)); dp[0] = 0; for(int i=1;i<=amount;i++) { for(int j=0;j<coins.size();j++)...