标签：ons bsp delete index examples ati 改进 rac lis
Given a string with no duplicate characters, return a list with all permutations of the characters.
Examples
time: O(n!), space: O(n)
public class Solution { public List<String> permutations(String set) { // Write your solution here List<String> res = new ArrayList<>(); if(set == null || set.length() == 0) { return res; } char[] chs = set.toCharArray(); boolean[] visited = new boolean[chs.length]; dfs(chs, visited, new StringBuilder(), res); return res; } private void dfs(char[] chs, boolean[] visited, StringBuilder sb, List<String> res) { if(sb.length() == chs.length) { res.add(sb.toString()); return; } for(int i = 0; i < chs.length; i++) { if(!visited[i]) { sb.append(chs[i]); visited[i] = true; dfs(chs, visited, sb, res); sb.deleteCharAt(sb.length() - 1); visited[i] = false; } } } }
改进：用swap，in-place操作，不需要StringBuilder
string长度为n，recursion level = n，第一层考虑位置0，将它和0,1,2,3,...n-1分别交换，有n种排法；第二层考虑位置1，10交换在第一层已经考虑过了，需要和剩下的1,2,3,4,...n-1交换，有n-1种排法；...；最后一层考虑位置n-1，只有一种排法，即和自身交换
time: O(n!), space: O(n)
public class Solution { public List<String> permutations(String set) { // Write your solution here List<String> res = new ArrayList<>(); if(set == null || set.length() == 0) { return res; } char[] chs = set.toCharArray(); dfs(chs, 0, res); return res; } private void dfs(char[] chs, int index, List<String> res) { if(index == chs.length) { res.add(new String(chs)); return; } for(int i = index; i < chs.length; i++) { swap(chs, index, i); dfs(chs, index + 1, res); swap(chs, index, i); } } private void swap(char[] arr, int i, int j) { char tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; } }
标签：ons bsp delete index examples ati 改进 rac lis
原文地址：https://www.cnblogs.com/fatttcat/p/10291879.html
