原题链接在这里：https://leetcode.com/problems/longest-uncommon-subsequence-ii/#/description
题目：
Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.
A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.
The input will be a list of strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn‘t exist, return -1.
Example 1:
Input: "aba", "cdc", "eae" Output: 3
Note:
题解：
对每一个string取出所有可能的substring 放到同一个map里计数. 找出计数为1的最长substring长度.
Time Complexity: O(n * 2^k). n = strs.length, k 是longest string的长度. 每一个s有2^s.length()个substring.
Space: O(k). stack space.
AC Java:
1 public class Solution { 2 public int findLUSlength(String[] strs) { 3 if(strs == null || strs.length == 0){ 4 return -1; 5 } 6 7 HashMap<String, Integer> hm = new HashMap<String, Integer>(); 8 for(String s : strs){ 9 for(String subStr : getSubsequences(s)){ 10 hm.put(subStr, hm.getOrDefault(subStr, 0)+1); 11 } 12 } 13 14 int res = -1; 15 for(Map.Entry<String, Integer> entry : hm.entrySet()){ 16 if(entry.getValue() == 1){ 17 res = Math.max(res, entry.getKey().length()); 18 } 19 } 20 return res; 21 } 22 23 private HashSet<String> getSubsequences(String s){ 24 HashSet<String> hs = new HashSet<String>(); 25 26 if(s.length() == 0){ 27 hs.add(""); 28 return hs; 29 } 30 31 HashSet<String> subHs = getSubsequences(s.substring(1)); 32 hs.addAll(subHs); 33 for(String str : subHs){ 34 hs.add(s.charAt(0)+str); 35 } 36 return hs; 37 } 38 }
类似Longest Uncommon Subsequence I.
LeetCode Longest Uncommon Subsequence II
原文：http://www.cnblogs.com/Dylan-Java-NYC/p/7052904.html