[Swift]LeetCode521.最长特殊序列Ⅰ|LongestUncommonSubsequenceI

OStack程序员社区-中国程序员成长平台 › 门户 › 编程› 移动开发›Swift 教程

原作者: [db:作者] 来自: [db:来源] 收藏邀请

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9812689.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. 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 two 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"
Output: 3
Explanation: The longest uncommon subsequence is "aba" (or "cdc"), 
because "aba" is a subsequence of "aba", 
but not a subsequence of any other strings in the group of two strings.

Note:

Both strings' lengths will not exceed 100.
Only letters from a ~ z will appear in input strings.

给定两个字符串，你需要从这两个字符串中找出最长的特殊序列。最长特殊序列定义如下：该序列为某字符串独有的最长子序列（即不能是其他字符串的子序列）。

子序列可以通过删去字符串中的某些字符实现，但不能改变剩余字符的相对顺序。空序列为所有字符串的子序列，任何字符串为其自身的子序列。

输入为两个字符串，输出最长特殊序列的长度。如果不存在，则返回 -1。

示例 :

输入: "aba", "cdc"
输出: 3
解析: 最长特殊序列可为 "aba" (或 "cdc")

说明:

两个字符串长度均小于100。
字符串中的字符仅含有 'a'~'z'。

8ms

 1 class Solution {
 2     func findLUSlength(_ a: String, _ b: String) -> Int {
 3         //每个字符串的最长子序列为其本身
 4         //最长特殊序列的长度必定为a.count 或者b.count或者-1,三者居其一。
 5         if a == b {return -1}
 6         else if a.count == b.count {return a.count}
 7         else
 8         {
 9             return max(a.count,b.count)
10         }
11     }
12 }

8ms

 1 import Foundation
 2 
 3 class Solution {
 4     func findLUSlength(_ a: String, _ b: String) -> Int {
 5         if (a == b) {
 6             return -1
 7         } else {
 8             return max(a.count,b.count)
 9         }
10     }
11 }

12ms

 1 class Solution {
 2     func findLUSlength(_ a: String, _ b: String) -> Int {
 3         let m = a.count
 4         let n = b.count
 5         
 6         if a == b {
 7             return -1
 8         }else if m > n {
 9             return m
10         }else{
11             return n
12         }
13     }
14 }

16ms

 1 class Solution {
 2     func findLUSlength(_ a: String, _ b: String) -> Int {
 3         guard a != b else {
 4             return -1
 5         }
 6         if a.count == 0 {
 7             return b.count
 8         }
 9         if b.count == 1 {
10             return a.count
11         }
12         let longer = a.count > b.count ? Array(a) : Array(b)
13         let shorter = a.count > b.count ? Array(b) : Array(a)
14         var count = 0
15         var start = 0
16         var end = longer.count
17         while start < end {
18             while end > start {
19                 if (String(shorter).range(of: String(longer[start..<end])) == nil) {
20                     count = max(longer[start..<end].count, count)
21                 }
22                 end -= 1
23             }
24             start += 1
25         }
26         return count
27     }
28 }