osigmao的专栏

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; }

two pointers, 两头往中间走

Remove系列：two pointers类型

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

//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]]++; }