題目：

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] ]

鏈接：?http://leetcode.com/problems/factor-combinations/

題解：

求一個(gè)數(shù)的所有factor，這里我們又想到了DFS + Backtracking， 需要注意的是，factor都是>= 2的，并且在此題里，這個(gè)數(shù)本身不能算作factor，所以我們有了當(dāng)n <= 1時(shí)的判斷 if(list.size() > 1) add the result to res.

Time Complexity - O(2ⁿ)， Space Complexity - O(n).

public class Solution {public List<List<Integer>> getFactors(int n) {List<List<Integer>> res = new ArrayList<>();List<Integer> list = new ArrayList<>();getFactors(res, list, n, 2);return res;}private void getFactors(List<List<Integer>> res, List<Integer> list, int n, int factor) {if(n <= 1) {if(list.size() > 1)res.add(new ArrayList<Integer>(list));return;}for(int i = factor; i <= n; i++) {if(n % i == 0) {list.add(i);getFactors(res, list, n / i, i);list.remove(list.size() - 1);}}} }

?

二刷：

還是使用了一刷的辦法，dfs + backtracking。但遞歸結(jié)束的條件更新成了n == 1。 但是速度并不是很快，原因是沒有做剪枝。我們其實(shí)可以設(shè)置一個(gè)upper limit，即當(dāng)i > Math.sqrt(n)的時(shí)候，我們不能繼續(xù)進(jìn)行下一輪遞歸，此時(shí)就要跳出了。

Java:

public class Solution {public List<List<Integer>> getFactors(int n) {List<List<Integer>> res = new ArrayList<>();if (n <= 1) return res;getFactors(res, new ArrayList<>(), n, 2);return res;}private void getFactors(List<List<Integer>> res, List<Integer> list, int n, int pos) {if (n == 1) {if (list.size() > 1) res.add(new ArrayList<>(list));return;}for (int i = pos; i <= n; i++) {if (n % i == 0) {list.add(i);getFactors(res, list, n / i, i);list.remove(list.size() - 1);}}} }

?

Update: 使用@yuhangjiang的方法，只用計(jì)算 2到sqrt(n)的這么多因子，大大提高了速度。

public class Solution {public List<List<Integer>> getFactors(int n) {List<List<Integer>> res = new ArrayList<>();if (n <= 1) return res;getFactors(res, new ArrayList<>(), n, 2);return res;}private void getFactors(List<List<Integer>> res, List<Integer> list, int n, int pos) {for (int i = pos; i <= Math.sqrt(n); i++) {if (n % i == 0 && n / i >= i) {list.add(i);list.add(n / i);res.add(new ArrayList<>(list));list.remove(list.size() - 1);getFactors(res, list, n / i, i);list.remove(list.size() - 1);}}} }

?

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Reference:

https://leetcode.com/discuss/51261/iterative-and-recursive-python

https://leetcode.com/discuss/87926/java-2ms-easy-to-understand-short-and-sweet

https://leetcode.com/discuss/58828/a-simple-java-solution

https://leetcode.com/discuss/72224/my-short-java-solution-which-is-easy-to-understand?

https://leetcode.com/discuss/82087/share-bit-the-thought-process-short-java-bottom-and-top-down

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