problems:
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)
Solution:依然采用先排序 然后左右夹逼 时间复杂度O(n3) 同2sum
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { vector<vector<int>> result; if(nums.size()<4) return result; sort(nums.begin(),nums.end()); auto last=nums.end(); for(auto a=nums.begin();a<prev(last,3);++a) for(auto b=next(a);b<prev(last,2);++b) { auto c=next(b); auto d=prev(last); while(c<d) { if(*a+*b+*c+*d<target) ++c; else if(*a+*b+*c+*d>target) --d; else{ result.push_back({*a,*b,*c,*d}); ++c; --d; } } } sort(result.begin(),result.end()); result.erase(unique(result.begin(),result.end()),result.end()); return result; } };
尽可能的降低时间复杂度,我们可以用一个hashmap来缓存两个数的和,这样子时间复杂度可以降低到O(n2)