Numbers can be regarded as product of its factors. For example,

8 = 2 x 2 x 2;
= 2 x 4.
Write a function that takes an integer n and return all possible combinations of its factors.

Note:

You may assume that n is always positive.
Factors should be greater than 1 and less than n.
Example 1:

Input: 1
Output: []
Example 2:

Input: 37
Output:[]
Example 3:

Input: 12
Output:
[
[2, 6],
[2, 2, 3],
[3, 4]
]
Example 4:

Input: 32
Output:
[
[2, 16],
[2, 2, 8],
[2, 2, 2, 4],
[2, 2, 2, 2, 2],
[2, 4, 4],
[4, 8]
]

我的答案：
本以為很簡單的一道題，沒想到竟然有這么兩個巨坑，不虧是M

• 坑1：for循環(huán)里面的結(jié)尾判斷是<還是<=
  • 我一開始想當(dāng)然覺得是<，因為從2開始，那么不可能搜索到n/2了，結(jié)果就被4打臉了，所以必須是<=，但這樣就造成后面很多問題了
• 坑2：如何去重？
  • 無腦篩選是會重復(fù)的，所以必須有個start的數(shù)字，for循環(huán)里面每個i，都不可能再新加一個<i的元素
  • 所以要screen out掉2，但是2<=2<=2，所以得在for里面的if也判斷一次，必須讓n/i這個可能新加入的元素也要>=i

class Solution {
public:
    vector<vector<int>> getFactors(int n, int start=2) {
        vector<vector<int>> ans;
        
        for (int i=start; i<=n/start; ++i) {
            // cout << n << " " << i << " " << n%i << endl;
            if (n%i == 0 and n/i>=i) {
                ans.push_back({n/i, i});
                for (auto& sub_factor:getFactors(n/i, i)) {
                    sub_factor.push_back(i);
                    ans.push_back(sub_factor);
                }
            }
        }
        
        return ans;
    }
};

Runtime: 92 ms, faster than 39.83% of C++ online submissions for Factor Combinations.
Memory Usage: 7.4 MB, less than 17.80% of C++ online submissions for Factor Combinations.

稍微優(yōu)化了一下代碼，用一個reference叫temp的vector<int>表示已經(jīng)搜索過的

class Solution {
private:
    vector<vector<int>> ans;
    
public:
    vector<vector<int>> getFactors(int n) {
        vector<int> temp;
        helper(n, temp, 2);
        
        return ans;
    }
    
    void helper(int n, vector<int>& temp, int start) {
        for (int i=start; i<=n/start; ++i) {
            if (n%i == 0 and n/i >= i) {
                temp.insert(temp.end(), {i,n/i});
                ans.push_back(temp);
                temp.pop_back();
                
                helper(n/i, temp, i);
                temp.pop_back();
            }
        }
    }
};

Runtime: 60 ms, faster than 44.49% of C++ online submissions for Factor Combinations.
Memory Usage: 7 MB, less than 76.27% of C++ online submissions for Factor Combinations.

如何成為0ms？

• 把i<=n/i放到for循環(huán)的結(jié)束判斷里

class Solution {
private:
    vector<vector<int>> ans;
    
public:
    vector<vector<int>> getFactors(int n) {
        vector<int> temp;
        helper(n, temp, 2);
        
        return ans;
    }
    
    void helper(int n, vector<int>& temp, int start) {
        for (int i=start; i<=min(n/start,n/i); ++i) {
            if (n%i == 0) {
                temp.insert(temp.end(), {i,n/i});
                ans.push_back(temp);
                temp.pop_back();
                
                helper(n/i, temp, i);
                temp.pop_back();
            }
        }
    }
};