② classSolution:#@return a tuple, (index1, index2)deftwoSum(self, num, target):#sortsorted_num =sorted(num)#two pointsleft =0 right= len(num) - 1res=[]while(left <right): sum= sorted_num[left] +sorted_num[right]ifsum ==target:#find out indexbreak;elifsum >target: right-= ...
arr, partition the array into (contiguous) subarrays of lengthat mostk. After partitioning, each subarray has their values changed to become the maximum value of that subarray. Returnthe largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a32...
} }thrownewIllegalArgumentException("No two sum solution"); } } 2.速度更快的解法,HashMap Copy classSolution{publicint[]twoSum(int[] nums,inttarget){ Map<Integer,Integer> m =newHashMap<Integer,Integer>();for(inti =0; i < nums.length; i++) {intcomplement = target - nums[i];if(m...
Split Array Largest Sum - Leetcode 410 - Python 16:51 Regular Expression Matching - Dynamic Programming Top-Down Memoization - Leetcod 27:56 Perfect Squares - Dynamic Programming - Leetcode 279 - Python 15:12 Pascal's Triangle - Leetcode 118 - Python 08:41 Partition Equal Subset Sum...
Max/LeetCodegitee.com/ProgrammerMax/leet-code.git 1. 题目:生命游戏—— 289 根据百度百科, 生命游戏 ,简称为 生命 ,是英国数学家约翰·何顿·康威在1970 年发明的细胞自动机。 给定一个包含 m× n 个格子的面板,每一个格子都可以看成是一个细胞。每个细胞都具有一个初始状态: 1 即为活细胞(live)...
链接:https://leetcode-cn.com/problems/combination-sum-iii 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 2,解题思路 采用dfs递归+剪枝的方法,借助于堆栈来实现; sum记录当前的累加和,depth记录当前数组中元素数目,tem记录当前存储的元素(dfs的路径),ans最终答案; ...
Leetcode 410 Split Array Largest SumGiven an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to mini…
Leetcode学习(2)—— Two Sum II - Input array is sorted,Givenanarrayofintegersthatisalreadysortediner.ThefunctiontwoSumshouldreturnindicesofthetwonumberssuc
2208. 将数组和减半的最少操作次数有时候真琢磨不透这个题的难度是怎么确定的。三档难度不适合现在的环境了，改成多挡难度制吧 优先
对于一些了解HashMap这种数据结构的同学,很容易能想到利用HashMap来求解,也就是和LeetCode第一题TwoSum相同的解法 classSolution{publicint[]twoSum(int[]numbers,inttarget){Map<Integer,Integer>map=newHashMap<>();for(inti=0;i<numbers.length;i++){// map.put(numbers[i], i); //如果放在这里可能会...