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 ...
privateintres = Integer.MAX_VALUE; publicintcoinChange(int[] coins,intamount) { Arrays.sort(coins); coinChange(coins, amount, coins.length-1,0); returnres==Integer.MAX_VALUE ? -1: res; } // privatevoidcoinChange(int[] coins,intamount,intcur,intcount) { if(cur <0) return;// stop...
dp初始状态:dp[0]=0,即当amount为0,以为coin的面值都是大于0的,所以不可能有任何任何一种硬币组合的面值为0 具体代码如下: class Solution: def coinChange(self, coins: List[int], amount: int) -> int: n, m = len(coins), amount dp = [0 for i in range(m + 1)] for i in range(1,...
Thought Process This is a classic problem and greedy might be the first thing comes into mind. For example, sort the coins on denomination and always use the largest amount coin. However, the optimal solution might not include the largest denomination. for example , [1, 3, 4] and target ...
Leetcode不定期更Up,深度学习NLP方向苦难研究生,人生体验派。人生得意须尽欢( ´ ▽ ` )ノ 746 直接 随意生成表情包 在线stable diffusion Liblib AI 下来播放 自动连播 :49 code力扣70. ClimbingStairs 爬楼梯(python版解析) 少女马曰曰 0 ...
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
1 2 3 4 5 6 classSolution{ public: intcoinChange(vector<int>&coins,intamount) { } }; 已存储 行1,列 1 运行和提交代码需要登录 coins = [1,2,5] amount = 11 9 1 2 3 4 5 6 › [1,2,5] 11 [2] 3 [1] 0 Source
LeetCode: 322. Coin Change 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 by any combination of the coins...
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...
计算并返回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。 你可以认为每种硬币的数量是无限的。 示例1: 输入:coins =[1, 2, 5], amount =11输出:3解释:11 = 5 + 5 + 1 示例2: 输入:coins =[2], amount =3输出:-1 ...