18. 4Sum

题目链接：https://leetcode.com/problems/4sum/

解题思路：

和3sum一样，不过固定前两个数而已。

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums.length<4 || nums==null)
            return res;
        HashSet<ArrayList<Integer>> hs = new HashSet<>();
        
        Arrays.sort(nums);
        
        for(int i=0;i<=nums.length-4;i++)
        {
            
            for(int j=i+1;j<=nums.length-3;j++)
            {
                int low = j+1;
                int high = nums.length-1;
                
                while(low < high)
                {
                    int sum = nums[i]+nums[j]+nums[low]+nums[high];
                    
                    if(sum==target)
                    {
                        ArrayList<Integer> un = new ArrayList<Integer>();
                        un.add(nums[i]);
                        un.add(nums[j]);
                        un.add(nums[low]);
                        un.add(nums[high]);
                        
                        if(!hs.contains(un))
                        {
                            res.add(un);
                            hs.add(un);
                        }
                        low++;
                        high--;
                    }
                    
                    else if(sum>target)
                    {
                        high--;
                    }
                    else if(sum<target)
                    {
                        low++;
                    }
                }
            }
            
           
        }
        return res;
    }
}