摘要： class MovingAverage { Queue<Integer> queue = new LinkedList<>(); int size = 0; double sum=0; public MovingAverage(int size) { this.size = size; } publ 阅读全文

摘要： class Solution { public boolean isPalindrome(String s) { s= s.toLowerCase(); int i=0, j=s.length()-1; while(i<j){ char a = s.charAt(i); char b = s.cha 阅读全文

摘要： We use two points to point to two successive nodes， we call them "pre" and "cur" We need to consider 3 situation need to be considered: 1. the inserte 阅读全文

摘要： The most important things to solve this problem are: recognizing this is a graph problem. Select a correct data structure to store the graph: Map<Stri 阅读全文

摘要： My PriorityQueue Solution: class Solution { public int[][] merge(int[][] intervals) { PriorityQueue<int[]> queue = new PriorityQueue<>((a,b)->a[0]-b[0 阅读全文

摘要： Just use BFS to solve this problem: 1. put the s to queue 2. if s is not a valid string, then remove a '(' or ')', and then put to the queue. 3. once 阅读全文

摘要： For this problem, we need to consider the following cases: +23 - valid 23+5 - invalid 9e - invalid e6 - invalid 23e+5 - valid 9.6e2 - valid 9e6.2 - in 阅读全文

摘要： My first solution - using ArrayList: class MyCircularDeque { List<Integer> list = new ArrayList<>(); int k ; public MyCircularDeque(int k) { this.k = 阅读全文

摘要： My first solution: class Solution { public int[] plusOne(int[] digits) { int n = digits.length; int carry = 1; for(int i=n-1;i>=0;i--){ int digit = di 阅读全文

摘要： Using binary search, time complexity: O(log(x)) class Solution { public int mySqrt(int x) { int l=1, r =x; while(l<r-1){ int mid=l+(r-l)/2; int temp = 阅读全文