【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：

 1 public class Solution {
 2     public static List<List<Integer>> fourSum(int[] nums, int target) {
 3         Arrays.sort(nums);
 4         //边界情况的判断
 5         List<List<Integer>> results = new ArrayList<List<Integer>>();
 6         if(nums.length<4){
 7             return results;
 8         }
 9         if(nums.length==4){
10             if(nums[0]+nums[1]+nums[2]+nums[3]==target){
11                 List<Integer> result = new ArrayList<Integer>();
12                 result.add(nums[0]);
13                 result.add(nums[1]);
14                 result.add(nums[2]);
15                 result.add(nums[3]);
16                 results.add(result);
17
18             }
19             return results;
20         }
21         //用遍历方式确定前两个数的位置
22         for(int first=0;first<nums.length-3;first++){
23             for(int second=first+1;second<nums.length-2;second++){
24                 //第三个数和第四个数用二分查找确定
25                 int third=second+1;
26                 int fourth=nums.length-1;
27                 while(third<fourth){
28                     if(nums[first]+nums[second]+nums[third]+nums[fourth]>target){
29                         fourth--;
30                     }
31                     else if(nums[first]+nums[second]+nums[third]+nums[fourth]<target){
32                         third++;
33                     }
34                     else{
35                         List<Integer> result = new ArrayList<Integer>();
36                         result.add(nums[first]);
37                         result.add(nums[second]);
38                         result.add(nums[third]);
39                         result.add(nums[fourth]);
40                         results.add(result);
41                         //避免元素重复
42                         while(third<nums.length-1&&nums[third+1]==nums[third]){
43                             third++;
44                         }
45                         third++;
46                         //避免元素重复
47                         while(third<fourth&&nums[fourth-1]==nums[fourth]){
48                             fourth--;
49                         }
50                         fourth--;
51
52                     }
53                 }
54                 //避免元素重复
55                 while(second<nums.length-2&&nums[second+1]==nums[second]){
56                     second++;
57                 }
58             }
59             //避免元素重复
60             while(first<nums.length-3&&nums[first+1]==nums[first]){
61                 first++;
62             }
63         }
64         return results;
65     }
66 }