18. 4Sum

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: 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]
]
代码解析：通过3Sum作为铺垫，在3sum中题目要求三个数相加为0, 4sum就可以理解为三个数相加为target-nums[k]只需要加一层循环去遍历这个数组即可，同样要做去重复这一步

public class Solution {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] nums = {-3,-2,-1,0,0,1,2,3};
        System.out.println(fourSum(nums, 0));
    }
     public static List<List<Integer>> fourSum(int[] nums,int target) {  
            List<List<Integer>> sumList = new ArrayList<List<Integer>>();  
            if(nums.length < 4)  
                return sumList;  
            Arrays.sort(nums);  
            for(int k = 0;k<nums.length;k++){  
                if (k == 0 || nums[k] > nums[k - 1]) {
            for(int i=k+1; i<nums.length-2; i++) {  

                if (i == 1 || nums[i] >= nums[i - 1]) {  
                int start = i+1;  
                int end = nums.length-1;  
                while(start<end) {  
                    if(nums[i]+nums[start]+nums[end] == target-nums[k]) {  
                        ArrayList<Integer> tempList = new ArrayList<Integer>(); 
                        tempList.add(nums[k]);
                        tempList.add(nums[i]);  
                        tempList.add(nums[start]);  
                        tempList.add(nums[end]); 
                        if(!sumList.contains(tempList))
                        sumList.add(tempList);  
                        start++;  
                        end--;  

                        //去掉重复的数据  
                        while((start<end) && (nums[end]==nums[end+1]))  
                            end--;  
                        while((start<end) && (nums[start]==nums[start-1]))  
                            start++;  
                    } else if(nums[i]+nums[start]+nums[end] > target-nums[k]) {  
                        end--;  
                    } else {  
                        start++;  
                    }  
                }  
            }  
            } 


                }
            }
            return sumList;  
        }  
}