LeetCode 15. 3Sum -- 数组中某三个元素之和为0，输出这三个元素的值，且这个三元组唯一

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Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

public class Solution {public List<List<Integer>> threeSum(int[] num) {        Arrays.sort(num);        List<List<Integer>> res = new LinkedList<>();//保证输出的顺序就是插入的顺序        for (int i = 0; i < num.length - 2; i++) {//比如：{-4, -1, -1, 0, 1, 5 ,5},最后一次比较的是1,5,5            if (i == 0 || (i - 1 >= 0 && num[i] != num[i - 1])) {//比如-1只能做基准1次，后面的-1直接跳过                int low = i + 1;                int high = num.length - 1;//最后一次num[i] = 1,num[low] = 5,num[high] = 5                while (low < high) {                    if (num[i] + num[low] + num[high] == 0) {                        res.add(Arrays.asList(num[i], num[low], num[high]));                        while (low < high && num[low] == num[low + 1]) low++;//比如：{-4, -1, -1, 0, 1, 5 ,5}，-1只能计算一次，不然会有重复                        while (low < high && num[high] == num[high - 1]) high--;//比如：{-4, -1, -1, 0, 1, 5, 5}，5只能计算一次，不然会有重复                        low++;                        high--;                    } else if (num[i] + num[low] + num[high] < 0) {                        low++;                    } else                        high--;                }            }        }        return res;    }//threeSum}

313 / 313 test cases passed.
Status: Accepted
Runtime: 77 ms
Your runtime beats 87.59 % of java submissions.

T(n) = O(n^2)

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