[LeetCode]018-4-Sum的c++解法：只能用哈希表和二叉树搜索方法

Question

求大神给下c++版的完整解法，包括头文件，主函数，类体。
请在文中注明一下具体而详细的思路方法，我学了哈希表和二叉树，一直都不太懂，感觉怪怪的。想要看看大神完整连贯的思路，模仿一下。c++解法，求大神自己敲一下，不要复制粘贴，菜鸡伤不起。

Answer 1

vector> fourSum(vector& nums, int target)
{
int n = nums.size();
QuickSort(nums,0,n-1);

    int i,j,k,s,sum,t1,t2;
    vector<int> turple;
    vector<vector<int>> result;
    if(n < 4)
        return result;
    t1 = nums[0];
    for(i=0;i<n-3;i++)
    {
        if(i && t1 == nums[i])
            continue;
        t2 = nums[i+1];
        for(j=i+1;j<n-2;j++)
        {
            if(t2 == nums[j] && j >i+1)
                continue;
            sum = target - nums[i] - nums[j];
            k = j+1;
            s = n-1;
            while(k<s)
            {
                if(nums[k] + nums[s] == sum)
                {
                    turple.push_back(nums[i]);
                    turple.push_back(nums[j]);
                    turple.push_back(nums[k]);
                    turple.push_back(nums[s]);
                    result.push_back(turple);
                    turple.clear();
                    k++;
                    s--;
                }
                else if(nums[k] + nums[s] < sum)
                    k++;
                else
                    s--;
             }
            t2 = nums[j];
         }
        t1 = nums[i];
    }
    result.erase(unique(result.begin(),result.end()),result.end());
    return result;
}
void QuickSort(vector<int>&nums,int low,int high)
{
    if(low<high)
    {
        int pivot = nums[low];
        int first = low;
        int last = high;
        while(first<last)
        {
            while(nums[last] > pivot && last > first)
            {
                 --last;
            }
            nums[first] = nums[last];
            while(nums[first] <= pivot && last >first)
            {
                ++first;
            }
            nums[last] = nums[first];
        }
        nums[first] = pivot;
        QuickSort(nums,low,first-1);
        QuickSort(nums,first+1,high);
    }
    else
        return;
}