Leetcode - 3 Sum

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)

[分析] Two Sum的扩展版，排序输入数组，遍历数组求解。遍历 nums[k]时，找出下标 k 后面所有加和为 -nums[k]的pair，因为数组已排序，可使用两指针从两端往中间夹逼的技巧求解 two sum。注意到 k 只需遍历到相应数组元素为0时即可停止，因为三个正数的和不可能是0，另外遍历时需要跳过重复元素。

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (nums == null || nums.length < 3)
            return result;
        Arrays.sort(nums);
        int N = nums.length;
        for (int k = 0; k < N - 2 && nums[k] <= 0; k++) {
            if (k > 0 && nums[k] == nums[k - 1])
                continue;
            int i = k + 1, j = N - 1;
            while ( i < j) {
                int sum2 = nums[i] + nums[j];
                if (sum2 == -nums[k]) {
                    List<Integer> item = new ArrayList<Integer>();
                    item.add(nums[k]);
                    item.add(nums[i]);
                    item.add(nums[j]);
                    result.add(item);
                    while (++i < j && nums[i] == nums[i - 1]);
                    while (i <--j && nums[j] == nums[j + 1]);
                } else if (sum2 < -nums[k]) {
                    while (++i < j && nums[i] == nums[i - 1]);
                } else {
                    while (i <--j && nums[j] == nums[j + 1]);
                }
            }
        }
        return result;
    }
}