osigmao的专栏

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

最值问题的第二种思路，保存局部最优，更新全局最优求P

找到存在的解：DFS + Backtracking因为ji

还要有多典型的DFS +  backtracking，之前写的乱七八糟凑出来的，后来用了这个si

1.标记 Visited:a. 直接在array中改 (bi'r

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

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 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

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,

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

two pointers, 两头往中间走

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

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