LeetCode : 3Sum [java]

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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:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
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)
思路：首先从小到大排序，第一层循环确定第一个数，相同的只取一次(第一次出现那个位置的)，然后，寻找第2个和第3个数，保证第2个数比第3个数小，使用两个指针从两侧进行扫描，发现和为0即记录，直到指针相遇，改变第一个数，继续扫描。
import java.util.ArrayList;import java.util.Arrays;import java.util.List;public class Solution {    public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> list = new ArrayList<List<Integer>>();if (nums.length < 3) {return list;}Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {if (i >= 1 && nums[i] == nums[i - 1]) {continue;}int j = i + 1;int k = nums.length - 1;while (j < k) {int sum = nums[i] + nums[j] + nums[k];if (sum > 0) {while (k > j && k <= nums.length - 2 && nums[k] == nums[k - 1]) {k--;}k--;} else if (sum < 0) {while (j < k && j >= 1 && nums[j] == nums[j + 1]) {j++;}j++;} else {List<Integer> interList = new ArrayList<>();interList.add(nums[i]);interList.add(nums[j]);interList.add(nums[k]);list.add(interList);while (j < k && j >= 1 && nums[j] == nums[j + 1]) {j++;}j++;}}}return list;    }}

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