标签：nat ase put == ref problem ++ dfs href
https://leetcode.com/problems/factor-combinations/description/
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:
Examples:
input: 1
output:
[]input:
37[]input:
12[ [2, 6], [2, 2, 3], [3, 4] ]input:
32[ [2, 16], [2, 2, 8], [2, 2, 2, 4], [2, 2, 2, 2, 2], [2, 4, 4], [4, 8] ]
Sol:
public class Solution { public List<List<Integer>> getFactors(int n) { // recursive DFS List<List<Integer>> result = new ArrayList<List<Integer>>(); helper(result, new ArrayList<Integer>(), n, 2); return result; } public void helper(List<List<Integer>> result, List<Integer> item, int n, int start){ // base case if (n <= 1){ if (item.size() > 1){ result.add(new ArrayList<Integer>(item)); } return ; } for (int i = start; i <= n; ++i){ if (n % i == 0){ item.add(i); helper(result, item, n/i, i); item.remove(item.size() - 1); } } } }
标签：nat ase put == ref problem ++ dfs href
原文地址：http://www.cnblogs.com/prmlab/p/7295264.html
