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:

  1. All the given strings' lengths will not exceed 10.
  2. The length of the given list will be in the range of [2, 50].

Approach #1: Simulate. [Java]

class Solution {

    public int findLUSlength(String[] strs) {
Arrays.sort(strs, new Comparator<String>() {
public int compare(String o1, String o2) {
return o2.length() - o1.length();
}
}); Set<String> duplicates = getDuplicates(strs);
for (int i = 0; i < strs.length; ++i) {
if (!duplicates.contains(strs[i])) {
if (i == 0) return strs[0].length();
for (int j = 0; j < i; ++j) {
if (isSubsequence(strs[j], strs[i])) break;
if (j == i - 1) return strs[i].length();
}
}
} return -1;
} boolean isSubsequence(String a, String b) {
int i = 0, j = 0;
while (i < a.length() && j < b.length()) {
if (a.charAt(i) == b.charAt(j)) ++j;
++i;
}
return j == b.length();
} Set<String> getDuplicates(String[] strs) {
Set<String> set = new HashSet<String>();
Set<String> duplicates = new HashSet<String>();
for (String str : strs) {
if (set.contains(str)) duplicates.add(str);
set.add(str);
}
return duplicates;
}
}

  

Analysis:

Sort the string in the reverse order. If there is not duplicates in the array, then the longest string is the answer.

But if there are duplicates, and if the longset string is not the answer, then we need to check other strings. But the smaller string can be subsequence of the bigger string. For this reason, we need to check if the string is a subsquence of all the strings bigger than itself. If not, that is the answer.

Reference:

https://leetcode.com/problems/longest-uncommon-subsequence-ii/discuss/99443/Java(15ms)-Sort-%2B-check-subsequence

522. Longest Uncommon Subsequence II的更多相关文章

  1. 【LeetCode】522. Longest Uncommon Subsequence II 解题报告(Python)

    [LeetCode]522. Longest Uncommon Subsequence II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemin ...

  2. 【leetcode】522. Longest Uncommon Subsequence II

    题目如下: 解题思路:因为given list长度最多是50,我的解法就比较随意了,直接用一个嵌套的循环,判断数组中每个元素是否是其他的subsequence,最后找出不属于任何元素subsequen ...

  3. 522 Longest Uncommon Subsequence II 最长特殊序列 II

    详见:https://leetcode.com/problems/longest-uncommon-subsequence-ii/description/ C++: 方法一: class Soluti ...

  4. [LeetCode] Longest Uncommon Subsequence II 最长非共同子序列之二

    Given a list of strings, you need to find the longest uncommon subsequence among them. The longest u ...

  5. [Swift]LeetCode522. 最长特殊序列 II | Longest Uncommon Subsequence II

    Given a list of strings, you need to find the longest uncommon subsequence among them. The longest u ...

  6. LeetCode Longest Uncommon Subsequence II

    原题链接在这里:https://leetcode.com/problems/longest-uncommon-subsequence-ii/#/description 题目: Given a list ...

  7. 【LeetCode Weekly Contest 26 Q2】Longest Uncommon Subsequence II

    [题目链接]:https://leetcode.com/contest/leetcode-weekly-contest-26/problems/longest-uncommon-subsequence ...

  8. Leetcode Longest Uncommon Subsequence I

    原题链接在这里:https://leetcode.com/problems/longest-uncommon-subsequence-i/#/description 题目: Given a group ...

  9. Longest Uncommon Subsequence I

    Given a group of two strings, you need to find the longest uncommon subsequence of this group of two ...

随机推荐

  1. Bootstrap学习遇到的role属性--- 无障碍网页应用属性

    以前接触过Bootstrap,但也只是仅仅接触,现在重新学习下,今天看到一个例子中的属性有一个role, 查阅资料发现这个是--WAI-ARIA无障碍设计属性: 通俗说是该设计为了一些盲人,失聪,残疾 ...

  2. random库的常见用法

    import random print( random.randint(1,10) ) # 产生 1 到 10 的一个整数型随机数 print( random.random() ) # 产生 0 到 ...

  3. 【转】手动释放linux os buff/cache

    手动释放linux内存cache和脚本定时释放 标签: linuxcache脚本bufferwindows磁盘 2011-12-04 08:44 12799人阅读 评论(2) 收藏 举报  分类: l ...

  4. mysql 表锁进程非常多的情况

    今天要说的是mysql 的 MYISAM引擎下的表锁问题. 通常来说,在MyISAM里读写操作是串行的,但当对同一个表进行查询和插入操作时,为了降低锁竞争的频率,根据concurrent_insert ...

  5. 2018.06.27The Windy's(费用流)

    The Windy's Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 6003 Accepted: 2484 Descripti ...

  6. Javascript php 异常捕获

    JavaScript try 语句允许我们定义在执行时进行错误测试的代码块. catch 语句允许我们定义当 try 代码块发生错误时,所执行的代码块. JavaScript 语句 try 和 cat ...

  7. springboot+websocket示例

    1.新建maven工程 工程结构如下: 完整的pom.xml如下: <?xml version="1.0" encoding="UTF-8"?> & ...

  8. INtellJ IDEA 2017 创建Annotation注解类

    1.建立一个文件夹 java ------->new ---->Package--->输入名字 2.New ---->java class --->如图修改红圈位置的下拉 ...

  9. VGA的行场时序

    之前碰到接收VGA时有的电脑可以有的电脑会出现画面偏移. 先来了解下数字显示器时序(DMT) DMT视频时序有四种: (1)Positive H & Positive V Syncs 行同步为 ...

  10. load data妙用

    load变量和用户变量的巧妙结合,实现灵活导入字段列(NO.1) LOAD DATA INFILE 'file.csv' INTO TABLE dados_meteo (@var1, @var2) S ...