1. 4Sum Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicate quadruplets. Fo…
描述 Follow up for "Remove Duplicates": What if duplicates are allowed at most twice? For example, Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3] 解法一 要解该题,只需要在上一题(leetcode解题报告(1):Remove Dupli…
LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Linked List Cycle II Given a linked list, return the node w…
最近在准备找工作的算法题,刷刷LeetCode,以下是我的解题报告索引,每一题几乎都有详细的说明,供各位码农参考.根据我自己做的进度持续更新中...... 本文地址 LeetCode:Reverse Words in a String LeetCode:Evaluate Reverse Polish Notation LeetCode:Max Points on a Line LeetCode:Sort List LeetCode:Ins…
LeetCode第27题 Given an array nums and a value val, remove all instances of that value in-place and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. The…
27. Remove Element Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can be changed.…
1.题目 27. Remove Element——Easy Given an array nums and a value val, remove all instances of that value in-place and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1)…
描述 Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can be changed. It doesn't matt…
描述 Given a sorted linked list, delete all duplicates such that each element appear only once. For example, Given 1->1->2, return 1->2. Given 1->1->2->3->3, return 1->2->3. 分析 只需要从头结点开始遍历,判断当前结点元素是否和下一个元素相等,如果相等,则跳过下一结点,否则直接让当前结点…