[Leetcode] 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)

Solution:

public class Solution {
    public List<List<Integer>> threeSum(int[] num) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (num.length < 3)
            return result;
        Arrays.sort(num);
        for (int i = 0; i < num.length - 2; ++i) {
            if (i > 0 && num[i] == num[i - 1])
                continue;
            int a = num[i];
            int low = i + 1;
            int high = num.length - 1;
            while (low < high) {
                if (-a == num[low] + num[high]) {
                    List<Integer> temp = new ArrayList<Integer>();
                    temp.add(a);
                    temp.add(num[low]);
                    temp.add(num[high]);
                    result.add(temp);
                    low++;
                    high--;
                    while (low < high && num[low] == num[low - 1])
                        low++;
                    while (low < high && num[high] == num[high + 1])
                        high--;
                } else if (-a > num[low] + num[high])
                    low++;
                else
                    high--;
            }
        }
        return result;
    }
}

注意：如果current value==previous value，需要跳过当前值。