osigmao的专栏

Sliding window类的题类似maximum subarray, maximum product subarray

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.Do not allocate extra space for another array, you must do this in place with

//二分搜索//考虑溢出的情况,所以要用long！要通过编译的话，还要强制类型转换回来！public class Solution {    public int sqrt(int x) {        if (x         long beg = 1;        long end = x/2;                while (beg   

1, recursion2, recursion + memoization3, DP

1. Inorder flatten:public class Solution { TreeNode prev = null; public void flatten(TreeNode root) { if (root == null) { return; } TreeNode rightnode =

1. x == reverse(x)?2. public class Solution { // 1. could negative number be Palindrome? // 2. reverse Integer! public boolean isPalindrome(int x) { if (x < 0) return false

Inorder逆向思维//inorderpublic class Solution { ListNode list; public TreeNode sortedListToBST(ListNode head) { int len = 0; ListNode node = head; while (node != null) {

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)You have the following 3 operations permitted on a word:

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a link

1.标记 Visited:a. 直接在array中改 (bi'r

Leetcode一些好的网站：1，九章算法da

DFS + BacktrackingII是考虑原set中有重复元素的情况Subset ISubset I

最值问题的第二种思路，保存局部最优，更新全局最优求P

public class Solution { public String multiply(String num1, String num2) { if (num1 == null || num2 == null) { return null; } int len1 = num1.length(); int l

1, 考虑溢出 -》用long。 If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.2，

先保存一下 某个大神的DP专辑：http://www.cnblogs.com/remlostime/tag/DP/

找到存在的解：DFS + Backtracking因为ji

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,

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.Above is a histogram where width o

array的找规律的题public class Solution { public int trap(int[] A) { int sum = 0; int maxInd = -1; int max = -1; //find the highest bars; for (int i = 0; i < A.leng

注意跳过重复的元素！//时间复杂度： O(n^2)public class Solution { public List> threeSum(int[] num) { List> res = new ArrayList<>(); if(num == null || num.length < 3) return res; Arra

two pointers O(n^2)public class Solution { public int threeSumClosest(int[] num, int target) { if (num == null || num.length <= 2) return 0; Arrays.sort(num); int closest

two pointers, 两头往中间走

Remove系列：two pointers类型

Linkedlist中的remove，记得要构造fakehead！

public class Solution { public List> fourSum(int[] num, int target) { List> res = new ArrayList<>(); if (num == null || num.length < 4) { return res; }

//法一： counting sortpublic class Solution { public void sortColors(int[] A) { int[] count = new int[3]; for (int i = 0; i < A.length; i++) { count[A[i]]++; }

public class Solution { public List> combinationSum(int[] candidates, int target) { if (candidates == null || candidates.length == 0) { return null; } Arrays.so

先把digit-letter建个Given a digit string, return all possible letter combinations that the number could represent.

public class Solution { public List grayCode(int n) { List res = new ArrayList<>(); res.add(0); if(n == 0) return res; res = grayCode(n-1); for(int

注意dividend是0的情况，这个时候sum总是比被除数大的！尼玛考虑负数的情况！而且还有溢出的情况。。。。 -（-2147483648）溢出啦 亲。。。要从O(n) -> O(lgn) -> O(1)

两次二分public class Solution { public int[] searchRange(int[] A, int target) { int[] res = new int[2]; res[0] = -1; res[1] = -1; int beg = 0; in

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

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

public class Solution { public String longestPalindrome(String s) { if (s == null) return null; int max = 0; String maxSub = ""; for (int beg = 0; beg < s.length();

枚举的题，一般用DFS+Backtracking做，此题会超时，那么要辅助memoization, 保存之前得到的结果.怎么mem？1，借鉴word break I, 保存能否该index之后能否break： boolean[s.length() + 1]2,  保存详细的分法结果： HashMap>public class Solution { List res =

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to