LeetCode 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)

 

类似3Sum，先排序，这后两重for循环，然后对最后的一个数组设两个指针遍历。这里注意重复的问题，例如第一重如果和前面一个数相同则跳过，因为前面的查找肯定包含了本次的情况。

O(n^3)

class Solution {
private:
    vector<vector<int> > ret;
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(num.begin(), num.end());
        
        ret.clear();
        
        for(int i = 0; i < num.size(); i++)
        {
            if (i > 0 && num[i] == num[i-1])
                continue;
                
            for(int j = i + 1; j < num.size(); j++)
            {
                if (j > i + 1 && num[j] == num[j-1])
                    continue;
                    
                int k = j + 1;
                int t = num.size() - 1;
                
                while(k < t)
                {
                    if (k > j + 1 && num[k] == num[k-1])
                    {
                        k++;
                        continue;
                    }
                    
                    if (t < num.size() - 1 && num[t] == num[t+1])
                    {
                        t--;
                        continue;
                    }
                    
                    int sum = num[i] + num[j] + num[k] + num[t];
                    
                    if (sum == target)
                    {
                        vector<int> a;
                        a.push_back(num[i]);
                        a.push_back(num[j]);
                        a.push_back(num[k]);
                        a.push_back(num[t]);
                        ret.push_back(a);
                        k++;
                    }
                    else if (sum < target)
                        k++;
                    else
                        t--;                        
                }
            }
        }
        
        return ret;
    }
};