打卡第五天～

题目描述

• 和这道题很相似，只是数据结构不同
  

思路 && 代码

• DFS + 辅助访问数组 + Set 减枝
• 通过 Set 实现去重

class Solution {Set<String> res = new HashSet<>();char[] arr;boolean[] visited;public String[] permutation(String s) {arr = s.toCharArray();visited = new boolean[arr.length];dfs(new StringBuilder());return res.toArray(new String[0]);}// 返回 s 的全排列（可重复）public void dfs(StringBuilder s) {// 找到一个结果的情况if(s.length() == arr.length) {res.add(s.toString());return;}Set<Character> set = new HashSet<>();for(int i = 0; i < arr.length; i++) {// 当前字符未访问if(!visited[i]) {visited[i] = true;// 当前字符第一次出现if(!set.contains(arr[i])) {dfs(s.append(arr[i]));set.add(arr[i]);s.delete(s.length() - 1, s.length());}// 恢复 visited[i]visited[i] = false;}}}
}

二刷

• 直觉没错，还是 DFS + Set 去重

class Solution {Set<String> set = new HashSet<>();boolean[] visited;char[] arr;public String[] permutation(String s) {visited = new boolean[s.length()];arr = s.toCharArray();dfs(new StringBuilder());return set.toArray(new String[0]);}void dfs(StringBuilder sb) {if(sb.length() == arr.length) {set.add(sb.toString());}for(int i = 0; i < arr.length; i++) {if(!visited[i]) {visited[i] = true;sb.append(arr[i]);dfs(sb);sb.deleteCharAt(sb.length() - 1);visited[i] = false;}}}
}