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LeetCode-18.4Sum

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums 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.

Example:

Given array nums = [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]
]

时间复杂度O(n³)

 1 public List<List<Integer>> fourSum(int[] nums, int target) {//set my
 2         List<List<Integer>> re = new ArrayList<>();
 3         //Set<List<Integer>> res = new HashSet<>();
 4         Arrays.sort(nums);
 5         for (int i = 0; i < nums.length; i++) {
 6             if(0<i&&nums[i]==nums[i-1]){
 7                 continue;
 8             }
 9             for (int j = i+1; j < nums.length; j++) {
10                 if((i+1)<j&&nums[j]==nums[j-1]){
11                     continue;
12                 }
13                 int k = j+1;
14                 int l = nums.length-1;
15                 while(k<l){
16                     if(target==(nums[i]+nums[j]+nums[k]+nums[l])){
17                         re.add(Arrays.asList(nums[i],nums[j],nums[k],nums[l]));
18                         while(k<l&&nums[k]==nums[k+1])k++;
19                         while(k<l && nums[l]==nums[l-1])l--;
20                         k++;
21                         l--;
22                     }
23                     else if(target>(nums[i]+nums[j]+nums[k]+nums[l])){
24                         k++;
25                     }
26                     else{
27                         l--;
28                     }
29                 }
30             }
31         }
32         return re;
33     }

该方法不是最优解

首先可以通过判断（可理解为剪枝）提高效率

是否存在低于O(n³)的方法？有待更新