[leetcode ] Factor Combinations -------------因数组合

Question:

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: 

1. Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not[6, 2].
2. You may assume that n is always positive.
3. Factors should be greater than 1 and less than n.

Examples: 
input: 1
output: 

[]

input: 37
output: 

[]

input: 12
output:

[  [2, 6],  [2, 2, 3],  [3, 4]]

input: 32
output:

[  [2, 16],  [2, 2, 8],  [2, 2, 2, 4],  [2, 2, 2, 2, 2],  [2, 4, 4],  [4, 8]]

分析：

题目中明确提出来，每个组合都要按照从小到大的顺序排列

采用回溯法，因数范围是2~n-1，每个组合的循环也是这个范围，第一次将n除以2，如果正好整除这将这两个数加入结果中，然后结果重复进行这个操作。然后回溯，回溯时候要注意将前一个除数删除。

代码如下：

<span style="font-size:14px;">class Solution {  public:      vector<vector<int>> getFactors(int n) {          vector<vector<int>> results;          vector<int> result;          getFactorsRecur(n, results, result, 0);          return results;      }            void getFactorsRecur(int n, vector<vector<int>> &results, vector<int> result, int last) {          for(int i = 2; i < n; i++) {              if(n / i >= i && i >= last && n % i == 0) {                  result.push_back(i);                  result.push_back(n / i);                  results.push_back(result);                  result.pop_back();                  getFactorsRecur(n / i, results, result, i);                  result.pop_back();              }          }      }  };  </span>