[LeetCode254]Factor Combinations

最新推荐文章于 2019-01-18 04:19:36 发布

SharonMoMo

最新推荐文章于 2019-01-18 04:19:36 发布

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本文介绍了一种算法，用于寻找一个正整数的所有可能因子组合，并确保每组因子按升序排列。通过两种方法实现：回溯法和优化后的递归搜索法。前者速度较慢但直观，后者利用了数学特性极大提高了效率。

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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: 
Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not [6, 2].
You may assume that n is always positive.
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]
]
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根据题目的描述，我们需要的factor starts from 2 end before n itself. 同时新factor要大于等于之前的factor。
所以如果n<=2 return [];
对于大于2的n，首先想到了backtracking 产生所有可能的combinations, 但是很慢188ms

    void helper(vector<vector<int>>& res, vector<int>& comb, int pos, int n){
        if(n<=1){
            if(comb.size() > 1){
                sort(comb.begin(), comb.end());
                res.push_back(comb);
            }
            return;
        }
        for(int i = pos; i<= n; ++i){
            if(!(n%i)){
                comb.push_back(i);
                helper(res, comb, i, n/i);
                comb.pop_back();
            }
        }
    }
    vector<vector<int>> getFactors(int n) {
        vector<vector<int>> res;
        vector<int> comb;
        if(n == 1) return res;
        helper(res, comb, 2, n);
        return res;
    }

另一种0ms的解法，终止条件是 i<=sqrt(n), 因为当i大于这个条件的时候，下一个factor必定小于当前的factor。所以code如下：

class Solution {
public:
    void helper(vector<vector<int>>& res, vector<int>& comb, int fac, int n){
        for(int i = fac; i<=sqrt(n); ++i){
            if(!(n%i)){
                vector<int> tmp = comb;
                tmp.push_back(i);//push current factor
                helper(res, tmp, i, n/i);//find all other factor smaller than sqrt(n/i);
                tmp.push_back(n/i);//add n/i(the largest factor under current combination.
                res.push_back(tmp);
            }
        }
    }
    vector<vector<int>> getFactors(int n){
        vector<vector<int>> res;
        vector<int> comb;
        helper(res, comb, 2, n);
        return res;
    }
};