标签：style blog http color 使用 os
Given a collection of numbers, return all possible permutations. For example, [1,2,3] have the following permutations: [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
Analysis: 这道题跟N-Queens，Sudoku Solver，Combination Sum，Combinations, Subsets, Generate Parentheses等一样，也是一个NP问题。
又是把所有满足条件的结果存到一个ArrayList里面， 之前也有类似的如Letter combination of a Phone Number这道题。这种类型的题其实形成了一个套路，套路就是，recursion参数包括最终结果的集合（ArrayList），input（String），递归层次level（int），某一条具体的路径Path
Notice: 我在permute函数里面第9行到第16行的循环其实是不需要的, 而且使用boolean数组来记录访问某个节点与否似乎比添加删除arraylist要容易
1 public class Solution { 2 public ArrayList<ArrayList<Integer>> permute(int[] num) { 3 ArrayList<ArrayList<Integer>> permutations = new ArrayList<ArrayList<Integer>>(); 4 ArrayList<Integer> permutation = new ArrayList<Integer>(); 5 ArrayList<Integer> collection = new ArrayList<Integer>(); 6 for (int each : num) { 7 collection.add(each); 8 } 9 for (int i = 0; i < collection.size(); i++) { 10 int current = collection.get(i); 11 collection.remove(i); 12 permutation.add(current); 13 helper(permutations, permutation, current, collection); 14 permutation.remove(0); 15 collection.add(i, current); 16 } 17 return permutations; 18 } 19 20 public void helper(ArrayList<ArrayList<Integer>> permutations, ArrayList<Integer> permutation, int current, ArrayList<Integer> collection) { 21 if (collection.isEmpty()) { 22 permutations.add(new ArrayList<Integer>(permutation)); 23 return; 24 } 25 for (int j = 0; j < collection.size(); j++) { 26 current = collection.get(j); 27 collection.remove(j); 28 permutation.add(current); 29 helper(permutations, permutation, current, collection); 30 permutation.remove(permutation.size()-1); 31 collection.add(j, current); 32 } 33 } 34 }
别人的做法：
public class Solution { public ArrayList<ArrayList<Integer>> permute(int[] num) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); ArrayList<Integer> element = new ArrayList<Integer>(); boolean[] visited = new boolean[num.length]; helper(num, result, element, visited); return result; } public void helper(int[] num, ArrayList<ArrayList<Integer>> result, ArrayList<Integer> element, boolean[] visited){ if (element.size() == num.length){ // duplicate element and add it to result (element would be changed from time to time. If directly use element // only result would be changed when element changed) result.add(new ArrayList<Integer>(element)); return; } for(int i=0; i<num.length; i++){ if(!visited[i]){ visited[i] = true; element.add(num[i]); helper(num, result, element, visited); // After providing a complete permutation, pull out the last number, element.remove(element.size()-1); visited[i] = false; } } } }
Leetcode: Permutations,布布扣,bubuko.com
标签：style blog http color 使用 os
原文地址：http://www.cnblogs.com/EdwardLiu/p/3811782.html
