[LeetCode]4Sum
4Sum
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)
四个的时候如果用3Sum的方法会超时。这时候应该想到二分搜索。确定两个值,然后二分搜索剩下的两个值。3Sum也可以用二分查找的方法,确定一个值,然后二分查找。
最后可以用set做一个去重的工作。
1 class Solution { 2 public: 3 vector<vector<int> > fourSum(vector<int> &nums, int target) { 4 vector<vector<int>> result; 5 if(nums.size()<4)return result; 6 sort(nums.begin(),nums.end()); 7 set<vector<int>> tmpres; 8 for(int i = 0; i < nums.size(); i++) 9 { 10 for(int j = i+1; j < nums.size(); j++) 11 { 12 int begin = j+1; 13 int end = nums.size()-1; 14 while(begin < end) 15 { 16 int sum = nums[i]+ nums[j] + nums[begin] + nums[end]; 17 if(sum == target) 18 { 19 vector<int> tmp; 20 tmp.push_back(nums[i]); 21 tmp.push_back(nums[j]); 22 tmp.push_back(nums[begin]); 23 tmp.push_back(nums[end]); 24 tmpres.insert(tmp); 25 begin++; 26 end--; 27 }else if(sum<target) 28 begin++; 29 else 30 end--; 31 } 32 } 33 } 34 set<vector<int>>::iterator it = tmpres.begin(); 35 for(; it != tmpres.end(); it++) 36 result.push_back(*it); 37 return result; 38 } 39 };