【leetcode】18. 4Sum

题目描述：

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.

解题分析：

这个和3Sum那道题有些像，也是要先确定两个数，之后两个数的确定可以参考二分查找的实现方法进行优化。

具体代码：

public class Solution {
    public static List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        //边界情况的判断
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        if(nums.length<4){
            return results;
        }
        if(nums.length==4){
            if(nums[0]+nums[1]+nums[2]+nums[3]==target){
                List<Integer> result = new ArrayList<Integer>(); 
                result.add(nums[0]);
                result.add(nums[1]);
                result.add(nums[2]);
                result.add(nums[3]);
                results.add(result);

            }
            return results;
        }
        //用遍历方式确定前两个数的位置
        for(int first=0;first<nums.length-3;first++){
            for(int second=first+1;second<nums.length-2;second++){
                //第三个数和第四个数用二分查找确定
                int third=second+1;
                int fourth=nums.length-1;
                while(third<fourth){
                    if(nums[first]+nums[second]+nums[third]+nums[fourth]>target){
                        fourth--;
                    }
                    else if(nums[first]+nums[second]+nums[third]+nums[fourth]<target){
                        third++;
                    }
                    else{
                        List<Integer> result = new ArrayList<Integer>();
                        result.add(nums[first]);
                        result.add(nums[second]);
                        result.add(nums[third]);
                        result.add(nums[fourth]);
                        results.add(result);
                        //避免元素重复
                        while(third<nums.length-1&&nums[third+1]==nums[third]){
                            third++;
                        }
                        third++;
                        //避免元素重复
                        while(third<fourth&&nums[fourth-1]==nums[fourth]){
                            fourth--;
                        }
                        fourth--;
                        
                    }
                }
                //避免元素重复
                while(second<nums.length-2&&nums[second+1]==nums[second]){
                    second++;
                }
            }
            //避免元素重复
            while(first<nums.length-3&&nums[first+1]==nums[first]){
                first++;
            }
        }
        return results;
    }
}