18. 4Sum leetcode (array) 总结

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: 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]
]

对与3sum,3sumclosest，4sum都采用了以下的思路方法：
(1)对数据进行排序
(2)遍历剩余两个数据，均采用二分查找进行逼近；
(3)4sum中点考虑了采用set进行去重

/*    对数组排序    确定四元数中的前两个（a，b）    遍历剩余数组确定两外两个（c，d），确定cd时思路跟3Sum确定后两个数据一样，二分查找左右逼近。    在去重时采用set集合    */    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> result;        set<vector<int>> setRes;        int len = nums.size();        sort(nums.begin(),nums.end());        if(len < 4)        {            return (vector<vector<int>> ());        }        /*        for(int i = 0; i < len; i++)        {            for(int j = i + 1; j < len; j++)        */        for(int i = 0; i < len - 3; i++)        {            for(int j = i + 1; j < len - 2; j++)            {                //二分查找                int begin = j + 1;                int end = len -1;                while(begin < end)                {                    int sum = nums[i] + nums[j] + nums[begin] + nums[end];                    if(sum == target)                    {                        vector<int> temp;                        temp.push_back(nums[i]);                        temp.push_back(nums[j]);                        temp.push_back(nums[begin]);                        temp.push_back(nums[end]);                        setRes.insert(temp);                        begin++;                        end--;                    }                    else if(sum > target)                    {                        end--;                    }                    else                    {                        begin++;                    }                }            }        }        set<vector<int>>::iterator iter;        for(iter = setRes.begin(); iter!=setRes.end(); iter++)        {            result.push_back(*iter);        }        return result;    }