3Sum

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)

题目大意：给定无序数组，找出和为0的不重复的三元组，三元组需以升序表示

题目难度：Medium

import java.util.*;

/**
 * Created by gzdaijie on 16/5/11
 * 先排序,后遍历
 * 遍历过程中,遇到每个元素,就检测其后面的每2个元素之和与之相加是否为0
 * 如何避免重复? 遍历下一个元素时,确保这个元素与上一个元素不同即可
 * 例如 -1,-1,-1,1,1,2,第一次是-1, 那么第二次就是 1
 * 时间复杂度O(N * lgN + N * N) => O(N * N)
 */
public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);

        int left, right, sum;
        ArrayList<Integer> triplets;

        for (int i = 0; i < nums.length - 2; i++) {
            if (i != 0) {
                // 为避免重复,i需要增加到与上次不同
                while (nums[i - 1] == nums[i] && i < nums.length - 2) ++i;
            }
            left = i + 1;
            right = nums.length - 1;

            while (left < right) {
                sum = nums[i] + nums[left] + nums[right];

                if (sum == 0) {
                    triplets = new ArrayList<>();
                    triplets.add(nums[i]);
                    triplets.add(nums[left]);
                    triplets.add(nums[right]);
                    result.add(triplets);
                    ++left;
                    // 避免后2个数重复,left需增加到与之前不同为止
                    while (left < right && nums[left - 1] == nums[left]) ++left;
                } else if (sum > 0) {
                    --right;
                } else {
                    ++left;
                }
            }
        }
        return result;
    }
}