[LeetCode]4Sum

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)

四个的时候如果用3Sum的方法会超时。这时候应该想到二分搜索。确定两个值，然后二分搜索剩下的两个值。3Sum也可以用二分查找的方法，确定一个值，然后二分查找。

最后可以用set做一个去重的工作。

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());
        set<vector<int>> tmpres;
        for(int i = 0; i < nums.size(); i++)
        {
            for(int j = i+1; j < nums.size(); j++)
            {
                int begin = j+1;
                int end = nums.size()-1;
                while(begin < end)
                {
                    int sum = nums[i]+ nums[j] + nums[begin] + nums[end];
                    if(sum == target)
                    {
                        vector<int> tmp;
                        tmp.push_back(nums[i]);
                        tmp.push_back(nums[j]);
                        tmp.push_back(nums[begin]);
                        tmp.push_back(nums[end]);
                        tmpres.insert(tmp);
                        begin++;
                        end--;
                    }else if(sum<target)
                        begin++;
                    else
                        end--;
                }
            }
        }
        set<vector<int>>::iterator it = tmpres.begin();
        for(; it != tmpres.end(); it++)
            result.push_back(*it);
        return result;
    }
};