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:
Note: 求和问题总结 http://tech-wonderland.net/blog/summary-of-ksum-problems.html
Use two pointers. First sort the array, and then from left to right, for each num[i], search the pair that sums up to -num[i] using Two Sum algorithm. 
Run time complexity is O(n^2), constant space.

public class Solution {
    public ArrayList> threeSum(int[] num) {
        ArrayList> result = new ArrayList>();
        Arrays.sort(num);
        
        if(num.length < 3)
            return result;
        
        for(int i = 0; i < num.length - 2; i++) {
            if(i == 0 || num[i] > num[i - 1]) {
                int x = -num[i];
                int start = i + 1;
                int end = num.length - 1;
                
                while(start < end) {
                    ArrayList temp = new ArrayList();
                    if(num[start] + num[end] == x) {
                        temp.add(num[i]);
                        temp.add(num[start]);
                        temp.add(num[end]);
                        result.add(temp);
                        start++;
                        end--;
                        
                        while(start < end && num[start] == num[start - 1]) {
                            start++;
                        }
                        while(start < end && num[end] == num[end + 1]) {
                            end --;
                        }
                    }else if(num[start] + num[end] > x) {
                        end--;
                    }else
                        start++;
                }
            }
        }
        return result;
    }
}