4Sum

题目

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)

思路一

类似于 3 Sum 的思路一或思路二，当然可以过小数据，但是对于大数据，还是Run Status: Time Limit Exceeded

 

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> result;  
        if(num.size()<4)  
            return result;  
        sort(num.begin(),num.end());      
        vector<int> myvec;  
        mythree(result,myvec,num,0,target);  
        return result;  
    }  
      
    void mythree(vector<vector<int>> &result, vector<int> &vec, vector<int> &num, int cur, int sum)  
    {  
        if(vec.size()==4)  
        {  
            if(sum==0)  
                result.push_back(vec);  
            return ;  
        }  
        if(vec.size()>4 || cur==num.size())  
            return ;       
        for(int i=cur;i<num.size();i++)  
        {  
            if(i>cur && num[i]==num[i-1])      
                continue;   
            vec.push_back(num[i]);  
            mythree(result,vec,num,i+1,sum-num[i]);  
            vec.pop_back();  
        }         
    }     
      
};  

思路二

用类似于3 Sum  思路三的方法，可以顺利的求解。

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = num.size();
        vector<vector<int> > ret;

        sort(num.begin(), num.end());
        for (int i = 0; i < n; ++i) {
            if (i != 0 && num[i] == num[i - 1]) continue;

            for (int j = n - 1; j > i; --j) {
                if (j != n - 1 && num[j] == num[j + 1]) continue;

                int L = i + 1, R = j - 1;
                while (L < R) {
                    int sum = num[i] + num[L] + num[R] + num[j];
                    if (L != i + 1 && num[L] == num[L - 1]) {
                        L++;
                    } else if (R != j - 1 && num[R] == num[R + 1]) {
                        R--;
                    } else if (sum > target) {
                        R--;
                    } else if (sum < target) {
                        L++;
                    } else {
                        ret.resize(ret.size() + 1);
                        ret.back().push_back(num[i]);
                        ret.back().push_back(num[L]);
                        ret.back().push_back(num[R]);
                        ret.back().push_back(num[j]);
                        L++;
                        R--;
                    }
                }
            }
        }
        return ret;
    }
};