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)

public class Solution {
    public List<List<Integer>> threeSum(int[] num) {
       List<List<Integer> > result = new ArrayList<List<Integer> >();
		Arrays.sort(num);
		int length = num.length;
		if(length >= 3) {
			for(int i = 0; i < length - 2 && num[i] <= 0; ++i) {
				int start = i + 1, end = length - 1;				
				//find all solution starting with num[i]
				while(start < end){
					int sum = num[i] + num[start] + num[end];
					if(sum < 0) //num[start] is too small
						++start;
					else if(sum > 0) //num[end] is too large
						--end;
					else { //find a solution
						List<Integer> oneSolution = new ArrayList<Integer>();
						oneSolution.add(num[i]);
						oneSolution.add(num[start]);
						oneSolution.add(num[end]);
						result.add(oneSolution);
						//remove the duplicates 
						do{
							++start;
						}while(start < end && num[start - 1] == num[start]);
						do{
							--end;
						}while(start < end && num[end] == num[end + 1]);
					}
				}
				//remove the duplicates of num[i] since we have found all solutions 
				//staring with num[i].
				while(i + 1 < length - 2 && num[i] == num[i + 1])
					++i;
			}
		}
		return result;
    }
}