ALiTuTu的专栏

看到要求的空间、时间复杂度完全没有头绪。只好看discussion好巧妙。。完全想不到。。。还是没有经验啊。。。。异或操作：相同为0，不同为1stem:Given an array of integers, every element appears twice except for one. Find that single one.Note:Your algorith

problem linkcode:public class Solution { public int findContentChildren(int[] g, int[] s) { Arrays.sort(g); Arrays.sort(s); int num=0; for(int i=0,

题目：https://leetcode.com/problems/add-digits/Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.For example:Given num = 38, the process is like: 3

problem linkword：in-place 在原位code:code1：计算某一非零位前面有多少个0，直接前移，然后给后面的赋值为0，有几个赋值几个。public class Solution { public void moveZeroes(int[] nums) { if(nums==null || nums.length==0)

problem linkcode：code1:递归在leetcode上要比BFS快一点public class Solution { public TreeNode invertTree(TreeNode root) { if(root == null) return null; TreeNode tmp = root.left;

problem linkcode1:将string变为char数组比第二种直接操作string使用charAt方法要快很多public class Solution { public char findTheDifference(String s, String t) { int sumS=0; int sumT=0; int

经典简单问题 problem linknote:（1）递归要有结束条件（2）此题可用DFS，BFS解题code：/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode r

problem linkcode:用负来标记出现过的，因为题目要求不能有extra space，复杂度为O(n)，所以只能操作原数组，另外可以利用的是数组的下标。public class Solution { public List findDisappearedNumbers(int[] nums) { for(int i=0;i<nums.length;

problem link:https://leetcode.com/problems/island-perimeter/code1:效率较高的两种解法，打败一半的人//查看右边和下边的邻居public class Solution { public int islandPerimeter(int[][] grid) { int islands = 0, ne

一个经典，基本的问题：翻转字符串problem link:https://leetcode.com/problems/reverse-string/note:(1)String类的方法substring(beginindex,endindex)两个参数如果相同返回的是空字符串(2)stringbuilder的replace方法和string的substring方法指针超

problem link：https://leetcode.com/problems/fizz-buzz/code：code1: 最简单的一种public class Solution { public List fizzBuzz(int n) { List res=new ArrayList(n); for(int i

problem link:https://leetcode.com/problems/hamming-distance/code:效率从高到低的三种方法第一种方法，要知道java中int占用4个字节（char占用2个字节 reference：http://blog.youkuaiyun.com/witsmakemen/article/details/8974200），这种类型的题既

Your runtime beats 82.86% of java submissions.stem：Write an algorithm to determine if a number is "happy".A happy number is a number defined by the following process: Starting with any pos

problem link看着复杂，其实转换一下思维就可以了code:public class Solution { //n-1个元素加一相当于只有一个减去1，那么每次只让一个元素减一，直到所有元素都相同，这个相同的元素就是数组中最小的元素 public int minMoves(int[] nums) { int min=nums[0];