Optimized Code: Use 2-pointer instead of StringBuilder to avoid extra space complexity class Solution { public boolean isPalindrome(String s) { int l = 0; int r = s.length() - 1; while (l < r) { char left = s.charAt(l); char right = s.charAt(r); if (!Character.isLetterOrDi...
publicclassSolution {publicintthreeSumClosest(int[] num,inttarget) {intresult = num[0] + num[1] + num[num.length - 1];//初始化Arrays.sort(num);//选择一个数for(inti = 0; i < num.length - 2; i++) {intstart = i + 1, end = num.length - 1;//双指针指向最小数和最大数wh...
1.两个指针i,j,当i<j时,不断两头往里面扫描 2.每扫描一个位置计算当前的面积,并刷新最大的面积 3.ai和aj的小者继续往里面递进 时间复杂度O(n) intmaxArea(int* height,intheightSize) {if(NULL==height)return0;inti=0,j=heightSize-1;intarea,t_max=0;while(i<j){ area=(j-i)*(height[i]...
技术标签:LeetCodeLinked ListTwo Pointer Description: Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: begin to intersect at node c1. Example 1: Exampl... ...
LeetCode Two Pointers related problems before I wrote this article, I just think that two pointer technique is nothing like iteration, but in an efficient way. it’s like brute force with memo. sometimes, I mean, many times, I just implement those things straightly, and after that I found...
two pointer的简单实例 time complexisty O(n), space complexisty O(1) two pointer跟binary search不要搞混了,binary search的每次计算量减半,所以time complexisty O(logn) Valid Palindrome https://leetcode.com/problems/valid-pali...leetcode Add Two Numbers Add Two Numbers python实现 方法一(自己实...
leetcode合并两个有序链表python 将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 示例: 输入:1->2->4, 1->3->4 输出:1->1->2->3->4->4 思路: 1、建立一个新链表存放排序后的节点 2、分别比较两个列表中的元素值,将小的元素节点添加到新的...
eg 7-10 Returns a pointer to the sum of two integers ...LeetCode-Sum of Two Integers Description: Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -. Example 1: Example 2: 题意:计算两个正数的和,要求不可以使用加法和减法; 解法:既然...
Question URL: https://leetcode.com/problems/two-sum/ Given an array of integersnumsand an integertarget, return indices of the two numbers such that they add up totarget. You may assume that each input would haveexactly one solution, and you may not use the same element twice. ...
This means if we find the intersection it will take at worst 2m + 2n pointer advancements. So O(m + n). They will never loop more than twice. If they don’t intersect then they reach null at the same time in the second pass because the pointers have been properly adjusted to end ...