通往梦想

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region . For example, X X X X X O O X X X O

Normal validation solution checking by row, column and square takes O(n^3) time, where n = 9. This solution takes O(n^2) time, but O(n^2) space. public boolean isValidSudoku(char[][] board) {

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combinatio

public class Solution { public boolean isValid(char[][] board, int x, int y){ int i, j; for(i = 0; i < 9; i++) if(i != x && board[i][y] == board[x][y]) return false; for(j = 0; j < 9;

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum

package Algorithm; import java.util.HashSet; import java.util.LinkedList; import java.util.Queue; public class WordLadder { public int ladderLength(String start, String end, HashSet dict)

Pay attention to this one. Passed after several bug fixes. public class Solution { public int threeSumClosest(int[] num, int target) { // Start typing your Java solution below // DO NOT w

public class Solution { public ArrayList> threeSum(int[] num) { // Start typing your Java solution below // DO NOT write main() function ArrayList> res = new ArrayList>(); if(num.length ==

public class Solution { public int[] searchRange(int[] A, int target) { // Start typing your Java solution below // DO NOT write main() function int[] res = {-1, -1}; if(A.length == 0)

public class Solution { public int search(int[] A, int target) { // Start typing your Java solution below // DO NOT write main() function if(A.length == 0) return -1; int begin = 0; i

import java.util.Hashtable; public class Solution { public int[] twoSum(int[] numbers, int target) { // Start typing your Java solution below // DO NOT write main() function int[] res = {0,

Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. Pay attenti

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr

This solution O(lg(m + n)) is not good for interview. Try kth smallest element solution extension. public class MedianTwoSortedArrays { private double findMedian(int A[], int B[], int left, i

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number

Same idea. public class Solution { public int counter; // 检测前面行可以放Q的位置是不是和目前位置（place[row]=j）冲突，即位置[i][place[i]]和位置[row][j]是否冲突 public boolean check(int row, int[] place) { for (int i = 0; i <

Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example, Given [100, 4, 200, 1, 3, 2], The longest consecutive elements sequence is [1, 2, 3,

Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order.The solution set must not contain duplica

public class Solution { public ArrayList res; public ArrayList generateParenthesis(int n) { // Start typing your Java solution below // DO NOT write main() function res = new ArrayList();

public class Solution { ArrayList res = null; public ArrayList letterCombinations(String digits) { // Start typing your Java solution below // DO NOT write main() function res = new A

public class Solution { public boolean isMatch(String s, String p) { // Start typing your Java solution below // DO NOT write main() function if (s.equals(p)) return true; else if (s.le

First solution (based on 3 sum). O(n^3). public class Solution { public ArrayList> fourSum(int[] num, int target) { // Start typing your Java solution below // DO NOT write main()

Given two integers n and k, return all possible combinations ofk numbers out of 1 ...n. For example, If n = 4 and k = 2, a solution is: [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]

Time Limit Exceeded, even for small judge. I guess it is because this solution needs n^2 space to store status for each Q placement. public class Solution { ArrayList res; public ArrayList

Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, [1,1,2] have the following unique permutations: [1,1,2], [1,2,1], and [2,1,1].

Given a collection of numbers, return all possible permutations. For example, [1,2,3] have the following permutations: [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. public class Solut

Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order.The solution set must not contain duplicate subsets. For example, I

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically n

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3): "123""132""213""231""312""

Kadane Algorithm O(n) Not working for all negative number arrays public class Solution { public int maxSubArray(int[] A) { // Start typing your Java solution below // DO NOT write main()