[leetcode] 18 4Sum

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问题描述：

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)

代码：

vector<vector<int> > fourSum(vector<int> &num, int target) {  //C++        vector<vector<int> > res;         if (num.size() <= 3) return res;        sort(num.begin(), num.end());        int twoSum;        for (int i = 0; i < num.size() - 3;)        {            for(int j = i+1; j < num.size()-2;){                int l = j+1, r = num.size() - 1;                twoSum = target -num[i] - num[j];                while (l < r)                {                    if (num[l] + num[r] < twoSum) l++;                    else if (num[l] + num[r] == twoSum)                    {                        vector<int> three(4);                        three[0] = num[i];                        three[1] = num[j];                        three[2] = num[l];                        three[3] = num[r];                        res.push_back(three);                        do { l++; }while (l < r && num[l - 1] == num[l]);                        do { r--; }while (l < r && num[r + 1] == num[r]);                    }                    else r--;                }                do{ j++; }while (j < num.size() - 2 && num[j - 1] == num[j]);            }            do{ i++; }while (i < num.size() - 3 && num[i - 1] == num[i]);        }        sort(res.begin(),res.end());        return res;    }

分析

此题的时间复杂度为O(n^3)。

对此题适当的变形：给定一个数组vec,问是否存在数组中的元素a,b,c,d 使得a+b+c+d =target。（a,b,c,d可以使同一个元素）

此题可以在O(n^2lgn)时间解决。

先对a+b的和进行穷举，保存在数组arr中。O（n^2）

然后对arr排序 O(n^2lgn)

两层循环遍历c，d ，二分查找是否c+d 存在于arr中。 O(n^2lgn)

所以总共的时间复杂度为O(n^2lgn)

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