[LeetCode]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)

4sum,可以归纳为3sum问题，这样复杂度 O(N^3)。但是实际中可以通过一些细节处理提高遍历无用解的情形。

16ms

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> ret;        int len = nums.size();        sort(nums.begin(),nums.end());        for(int i=0;i<len-3;++i){            if(i>0&&nums[i]==nums[i-1]) continue;//dup start            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;//no result            if(nums[i]+nums[len-1]+nums[len-2]+nums[len-3]<target) continue;//at this start no result,go to next one,important            for(int j=i+1;j<len-2;++j){//3 sum                if(j>i+1&&nums[j]==nums[j-1]) continue;//dup start                if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;//优化的关键                if(nums[i]+nums[j]+nums[len-1]+nums[len-2]<target) continue; //优化的关键                int left = j+1;                int right = len-1;                while(left<right){                    if(nums[i]+nums[j]+nums[left]+nums[right]==target){                        ret.push_back({nums[i],nums[j],nums[left],nums[right]});                        ++left;                        --right;                        while(nums[left]==nums[left-1]&&nums[right]==nums[right+1]){ //重复子解,重要                            ++left;                            --right;                        }                    }                    else if(nums[i]+nums[j]+nums[left]+nums[right]<target){                         ++left;                        while(nums[left]==nums[left-1]){//重复解，重要                            ++left;                        }                    }                    else if(nums[i]+nums[j]+nums[left]+nums[right]>target){                         --right;                        while(nums[right]==nums[right+1]){ //重复解                            --right;                         }                    }                }            }        }        return ret;    }};