LeetCode 46: Permutations

最新推荐文章于 2021-02-04 14:52:57 发布

转载最新推荐文章于 2021-02-04 14:52:57 发布 · 47 阅读

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原文链接：http://www.cnblogs.com/shuashuashua/p/7223514.html

本文提供了两种不同的排列算法实现：迭代法和递归法。迭代法通过逐层生成新的排列组合来解决问题，而递归法则利用深度优先搜索的方式进行。这两种方法都能有效生成给定数组的所有可能排列。

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Tips :

1. Need to new array list since it is passed by reference. All the values will be cleaned up the original data is used.

2. For iteraive, no need to copy the last loop result since it should be a new B+ tree layer generation.

Iterative :

 1 public class Solution {
 2     public List<List<Integer>> permute(int[] nums) {
 3         List<List<Integer>> result = new ArrayList<>();
 4         result.add(new ArrayList<>());
 5         
 6         for (int i = 0; i < nums.length; i++) {
 7             List<List<Integer>> newResult = new ArrayList<>();
 8             for (List<Integer> internal : result) {
 9                 for (int j = 0; j <= i; j++) {
10                     List<Integer> newInternal = new ArrayList<>(internal);
11                     newInternal.add(j, nums[i]);
12                     newResult.add(newInternal);
13                 }
14             }
15             result = newResult;
16         }
17         return result;
18     }
19 
20 }

Recursive :

 1 public class Solution {
 2     public List<List<Integer>> permute(int[] nums) {
 3         List<List<Integer>> result = new ArrayList<>();
 4         DFS(result, new ArrayList<>(), nums, new boolean[nums.length]);
 5         return result;
 6     }
 7     
 8     private void DFS(List<List<Integer>> result, List<Integer> currentList, int[] nums, boolean[] visited) {
 9         if (currentList.size() == nums.length) {
10             result.add(new ArrayList<>(currentList));
11             return;
12         }
13         
14         for (int i = 0; i < nums.length; i++) {
15             if (visited[i]) {
16                 continue;
17             }
18             visited[i] = true;
19             currentList.add(nums[i]);
20             DFS(result, currentList, nums, visited);
21             currentList.remove(currentList.size() - 1);
22             visited[i] = false;
23         }
24     }
25 }

转载于:https://www.cnblogs.com/shuashuashua/p/7223514.html