LeetCode 254. Factor Combinations（因式分解）

原题网址：https://leetcode.com/problems/factor-combinations/

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]
]

思路：深度优先搜索。

public class Solution {private void find(List<Integer> factors, int target, List<List<Integer>> results) {// System.out.printf("factors=%s\n", factors);if (target == 1) {if (!factors.isEmpty()) {List<Integer> result = new ArrayList<>(factors.size());result.addAll(factors);results.add(result);}return;}int min = factors.isEmpty()? 2 : factors.get(factors.size()-1);int max = factors.isEmpty()? target/2 : target;for(int f = min; f <= max; f ++) {if (target % f != 0) continue;factors.add(f);find(factors, target/f, results);factors.remove(factors.size()-1);}}public List<List<Integer>> getFactors(int n) {List<List<Integer>> results = new ArrayList<>();find(new ArrayList<>(), n, results);return results;}
}

优化：关键在于i*i<=n，可以大幅提升性能。

public class Solution {private void find(int from, int n, List<Integer> factors, List<List<Integer>> results) {for(int i=from; i*i<=n; i++) {if (n % i == 0) {factors.add(i);find(i, n/i, factors, results);factors.remove(factors.size()-1);}}if (!factors.isEmpty()) {factors.add(n);results.add(new ArrayList<>(factors));factors.remove(factors.size()-1);}}public List<List<Integer>> getFactors(int n) {List<List<Integer>> results = new ArrayList<>();find(2, n, new ArrayList<>(), results);return results;}
}

由于ArrayList较慢，如果固定分配空间可以进一步提升性能。

public class Solution {private void find(int from, int n, int[] factors, int len, List<List<Integer>> results) {for(int i=from; i*i<=n; i++) {if (n % i == 0) {factors[len] = i;find(i, n/i, factors, len+1, results);}}if (len > 0) {factors[len] = n;Integer[] f = new Integer[len+1];for(int i=0; i<=len; i++) f[i] = factors[i];results.add(Arrays.asList(f));}}public List<List<Integer>> getFactors(int n) {int p = 0;for(int i=1; i<=n; p++, i*=2);List<List<Integer>> results = new ArrayList<>();find(2, n, new int[p], 0, results);return results;}
}

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