LeetCode18 4Sum

题意：

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: The solution set must not contain duplicate quadruplets.(Medium)

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]
]

分析：

还是采取跟2sum,3sum一样的思路，先排序，再two pointers.

实现的时候利用3sum，然后加一层循环，复杂度O(n^3).

代码：

 class Solution {
 private:
     vector<vector<int>> threeSum(vector<int>& nums, int target) {
         vector<vector<int>> v;
         if (nums.size() < ) {  //数组元素个数过少，直接返回
             return v;
         }
         for (int i = ; i < nums.size() - ; ++i) {
             if (i >=  && nums[i] == nums[i - ]) {
                 continue;
             }
             int start = i + , end = nums.size() - ;
             while (start < end) {
                 if ( nums[i] + nums[start] + nums[end] == target ) {
                     vector<int> temp{nums[i], nums[start], nums[end]};
                     v.push_back(temp);
                     start++;
                     end--;
                     while ( (start < end) && nums[start] == nums[start - ]) { //没加start < end虽然过了，估计是样例不够完善
                         start++;
                     }
                     while ( (start < end) && nums[end] == nums[end + ]) {
                         end--;
                     }
                 }
                 else if (nums[i] +  nums[start] + nums[end] > target ) {
                     end--;
                 }
                 else {
                     start++;
                 }
             }
         }
         return v;
     }
 public:
     vector<vector<int>> fourSum(vector<int>& nums, int target) {
         sort(nums.begin(), nums.end());
         vector<vector<int>> v;
         for (int i = ; i < nums.size(); ++i) {
             if (i >  && nums[i] == nums[i - ]) {
                 continue;
             }
             vector<int> temp(nums.begin() + i + , nums.end());
             vector<vector<int>> result = threeSum(temp, target - nums[i]);
             for (int j = ; j < result.size(); ++j) {
                 result[j].push_back(nums[i]);
                 v.push_back(result[j]);
             }
         }
         return v;
     }
 };