[leetcode] 18 4Sum

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)