[抄题]:
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]
]
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
知道是backtracing可是动不了笔:扩展就是n % i = 0添加因数就行了,退出条件就是把item添加到结果中
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 递推表达式中的因数n要变成n/i
- 退出条件是n<= 1就肯定要用return退出,是否添加取决于item的size是否大于而不是等于1
[二刷]:
helper里面就是个参数,需要从start开始
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
- 退出条件是n<= 1就肯定要用return退出,是否添加取决于item的size是否大于而不是等于1
[复杂度]:Time complexity: O(乘以每个点是nlgn) Space complexity: O(递归树是lgn)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
class Solution {
public List<List<Integer>> getFactors(int n) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (n <= 0) return result;
helper(2, new ArrayList<Integer>(), n, result);
return result;
} public void helper(int start, List<Integer> item, int n, List<List<Integer>> result) {
//add the item to the result
if (n <= 1) {
if (item.size() > 1)
result.add(new ArrayList<Integer>(item));
return;
} //calculate the factors
for (int i = start; i <= n; i++) {
if (n % i == 0) {
item.add(i);
helper(i, item, n / i, result);
item.remove(item.size() - 1);
}
}
}
}