Leetcode and Lintcode

3Sum

Problem:

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)

Leetcode link
Lintcode link

My Solution:

Sort the list first, which takes O(Nlog N). Then go through the sorted list， for each integer “nums[i]” in the list,
narrow the range from the integers at index i and the end of the list. It takes O(n^2) 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
/*************************************************************************
	> File Name: ThreeSum.java
	> Author: Yao Zhang 
	> Mail: psyyz10@163.com 
	> Created Time: Wed 11 Nov 18:17:20 2015
 ************************************************************************/

public class ThreeSum{
    public List<List<Integer>> threeSum(int[] nums){
        List<List<Integer>> result = new ArrayList<List<Integer>>();
    
        if (nums.length < 3)
            return result;
                     
        Arrays.sort(nums);
    
        for (int i = 0; i < nums.length; i++){
            if (i != 0 && nums[i] == nums[i - 1]) 
                continue;
                     
            int start = i + 1;
            int end = nums.length - 1;
    
            while (start < end){
                if (nums[i] + nums[start] + nums[end] == 0){
                    List<Integer> list = new ArrayList<Integer>();
                    list.add(nums[i]);
                    list.add(nums[start]);
                    list.add(nums[end]);
                    result.add(list);
    
                    start++;
                    end--;
    
                    while (nums[start - 1] == nums[start] && start < end)
                        start++;
    
                    while(nums[end + 1] == nums[end] && start < end)
                        end--;
                } else if (nums[i] + nums[start] + nums[end] < 0){
                    start++;
                } else{ 
                    end--;
                }    
            }        
        }            
        return result;    
    }
}

Related Problem:
Two Sum
4Sum
3Sum Closest