LeetCode OJ：4Sum（4数字之和）

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.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {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)

和前面的思路基本上是相同的，这里同样容易遇见相同的数，这个时候需要跳过，代码如下：

 class Solution {
 public:
     vector<vector<int>> fourSum(vector<int>& nums, int target) {
         sort(nums.begin(), nums.end());
         int sz = nums.size();
         vector<vector<int>> ret;
         for(int i = ; i < sz; ++i){
             if(i !=  && nums[i] == nums[i - ])
                 continue;
             for(int j = i + ; j < sz; ++j){
                 if(j > i +  && nums[j] == nums[j - ])
                     continue;
                 for(int beg = j + , end = sz - ; beg < end;){
                     while(beg > j +  && nums[beg] == nums[beg - ])
                         beg++;
                     while(end < sz -  && nums[end] == nums[end + ])
                         end--;
                     if(beg >= end) break;
                     int sum = nums[i] + nums[j] + nums[beg] + nums[end];
                     if(sum == target){
                         vector<int> tmp;
                         tmp.push_back(nums[i]);
                         tmp.push_back(nums[j]);
                         tmp.push_back(nums[beg]);
                         tmp.push_back(nums[end]);
                         ret.push_back(tmp);
                         beg++, end--;
                     }else if(sum < target){
                         beg++;
                     }else{
                         end--;
                     }
                 }
             }
         }
         return ret;
     }
 };