Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters.

Return the maximum possible length of s.

Example 1:

Input: arr = ["un","iq","ue"]
Output: 4
Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
Maximum length is 4.

Example 2:

Input: arr = ["cha","r","act","ers"]
Output: 6
Explanation: Possible solutions are "chaers" and "acters".

Example 3:

Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
Output: 26

Constraints:

• 1 <= arr.length <= 16
• 1 <= arr[i].length <= 26
• arr[i] contains only lower case English letters.

串联字符串的最大长度。

给定一个字符串数组 arr，字符串 s 是将 arr 某一子序列字符串连接所得的字符串，如果 s 中的每一个字符都只出现过一次，那么它就是一个可行解。

请返回所有可行解 s 中最长长度。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters
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思路是DFS backtracking。这里我用一个 letters 数组判断当前拼凑出来的 path 里面是否有重复字母，其他部分基本是标准的DFS。

时间O(n*2^n)

空间O(n)

Java实现

 1 class Solution {
 2     int res = 0;
 3 
 4     public int maxLength(List<String> arr) {
 5         // corner case
 6         if (arr == null || arr.size() == 0) {
 7             return 0;
 8         }
 9         dfs(arr, "", 0);
10         return res;
11     }
12 
13     private void dfs(List<String> arr, String path, int index) {
14         boolean isUnique = helper(path);
15         if (isUnique) {
16             res = Math.max(res, path.length());
17         }
18         if (index == arr.size() || !isUnique) {
19             return;
20         }
21         for (int i = index; i < arr.size(); i++) {
22             dfs(arr, path + arr.get(i), i + 1);
23         }
24     }
25 
26     private boolean helper(String s) {
27         boolean[] letters = new boolean[26];
28         for (int i = 0; i < s.length(); i++) {
29             if (letters[s.charAt(i) - 'a'] == true) {
30                 return false;
31             }
32             letters[s.charAt(i) - 'a'] = true;
33         }
34         return true;
35     }
36 }

 

LeetCode 题目总结