Leetcode and Lintcode

4Sum

Problem:

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)

Leetcode link
Lintcode link

My solution:

The solution is nearly the same as 3sum

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/*************************************************************************
	> File Name: FourSum.java
	> Author: Yao Zhang 
	> Mail: psyyz10@163.com 
	> Created Time: Wed 11 Nov 23:48:43 2015
 ************************************************************************/

public class FourSum{
    public ArrayList<ArrayList<Integer>> fourSum(int[] nums, int target) {     
        //write your code here
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (nums.length < 4) return result;
        Arrays.sort(nums);                 
         
        // note i < nums.length - 3                            
        for (int i = 0; i < nums.length - 3; i ++){
            if (i != 0 && nums[i - 1] == nums[i])
                continue;                  
            
            // note j start from i + 1, but not i                               
            for (int j = i + 1; j < nums.length - 2; j++){
                if (j != i + 1 && nums[j - 1] == nums[j])
                    continue;              
                                           
                int start = j + 1;         
                int end = nums.length - 1;
                                           
                while (start < end){       
                    int sum = nums[i] + nums[j] + nums[start] + nums[end];
                    if (sum == target){
                       ArrayList<Integer> list = new ArrayList<Integer>();
                       list.add(nums[i]);
                       list.add(nums[j]);
                       list.add(nums[start]);
                       list.add(nums[end]);
                       result.add(list);                                                                                                                                             
                       start++;            
                       end--;              
                                           
                      // note condition start < end
                       while (nums[start - 1] == nums[start] && start < end) 
                           start++;        
                                           
                       while (nums[end + 1] == nums[end] && start < end)
                           end--;          
                    } else if (sum < target){
                        start++;           
                    } else {               
                        end--;             
                    }                      
                }                          
            }                              
        }                                  
                                           
        return result; 
    }
}

Related Problem:
Two Sum
3Sum
3Sum Closest