18. 4Sum（双指针 注意答案去重）

---

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]
]

 

 1 class Solution {
 2 public:
 3     vector<vector<int>> fourSum(vector<int>& nums, int target) {
 4         vector<vector<int>> res;
 5         if(nums.size() < 4) return res;
 6         sort(nums.begin(),nums.end());
 7         for(int i = 0;i < nums.size() - 3;++i) {
 8             if (i!=0 && nums[i]==nums[i-1]) continue; // 去重
 9             for(int j = i+1;j < nums.size() -2;++j) {
10                 if(j!= i+1 && nums[j]==nums[j-1]) continue;//去重
11                 int low = j + 1;
12                 int high = nums.size() -1;
13                 while(low < high) {
14                     int sum = nums[i] + nums[j] + nums[low] + nums[high];
15                     if (sum == target) {
16                         res.emplace_back(vector<int>({nums[i],nums[j],nums[low],nums[high]}));
17                         while(low < high && nums[low]==nums[low+1]) low++;  // 去重
18                         while(low < high && nums[high]==nums[high-1]) high--; // 去重
19                         high--;
20                         low++;
21                     } else if (sum > target) {
22                         high--;
23                     } else if (sum < target) {
24                         low++;
25                     }
26                 }
27             }
28         }
29         return res;
30     }
31 };

 

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