4sum leetcode

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)

思路分析：

　　我采用的是改善的暴力破解！

　　首先当然是排序！

　　最外面用一个双循环，一个从前往后，一个从后往前遍历，并且完成一次遍历之后分别都要去重！

　　里面使用一个循环，2个指针同时进行，并且根据结果来移动两个指针，比如若是4个数的结果比target大，说明，应当将第二个指针往前移。也要进行去重！

　　代码如下：（参考了网上的资料，可以采用分治法，将所有两个数的组合的结果储存起来，然后再进行任意两个数的组合！！！后面再尝试写一下，估计会用到hash来记录对应的元素？）

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

class Solution {
private:
    vector<vector<int>> res;
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int len = nums.size();
    
        sort(nums.begin(), nums.end());

        for (int i = 0; i < nums.size(); ++i)
        {

            for (int j = nums.size() - 1; j>i; --j)
            {

                helper(nums, target, i, j);

                while (nums[j - 1] == nums[j])//去重
                    --j;
            }
            while (i<len-1&&nums[i + 1] == nums[i])//去重
            ++i;
        }
            return res;
    }
    void helper(vector<int>&nums, int target, int index1, int index2)
    {
        int sum;

        for (int i = index1 + 1,j=index2-1; i < j; )
        {
            sum = nums[index1] + nums[index2] + nums[i] + nums[j];//进行比较，这是比较常用的做法来一次完成里面的遍历
            if (sum == target)
            {
                vector<int> temp;
                temp.push_back(nums[index1]);
                temp.push_back(nums[i]);
                temp.push_back(nums[j]);
                temp.push_back(nums[index2]);
                res.push_back(temp);
                while (nums[i + 1] == nums[i] && i < j)//去重
                    i++;
                while (nums[j - 1] == nums[j] && i < j)//去重
                    --j;
                ++i;
                --j;
            }
            else if (sum < target)
                ++i;
            else --j;
        
        }
    }
};

int main()
{
    Solution test;
    vector<int> val = { -1,0,1,2,-1,-4};
    //for (int i = 0; i < val.size(); ++i)
    //    cout << val[i] << " ";
    //cout << endl;
    vector<vector<int>> result = test.fourSum(val, -1);
    for (int i = 0; i < result.size(); ++i)
    {

        for (int j = 0; j < result[i].size(); ++j)
            cout << result[i][j] << " ";
        cout << endl;
    }

    return 0;
}