LeetCode 521：Longest Uncommon Subsequence I （C++）

一：题目

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 beany 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: 3Explanation: 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:

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

二：解体分析

（百度翻译）

给定一组两个字符串，您需要找到这组字符串的最长不寻常子序列。最长不寻常子序列被定义为这些字符串之一的最长子序列，这个子序列不应该是其他字符串的任何子序列。            

子序列是一个序列，它可以从一个序列中派生出来，通过删除一些字符而不改变其余元素的顺序。显然，任何字符串本身是一个序列和一个空的字符串是一个序列的任何字符串。            

输入将是两个字符串，输出需要是最长不寻常子序列的长度。如果最长的异常子序列不存在，返回- 1。

一开始想的很复杂，看了网上的一些帖子发现，只需判断两个字符串是否相等，

如果相等，则不存在最长的不寻常子序列，返回-1；

如果不相等，则最长的字符串就是最长的不寻常子序列，因为在另一个字符串中可定找不到其子序列

三：代码实现

class Solution {public:    int findLUSlength(string a, string b) {        //如果字符a与b相同，则返回-1        //否则，输出最大的字符串长度        //刚看题时觉得很难，看到网友的分析后，发现是很简单的思路        if(a==b)            return -1;        else            return a.length()>b.length()?a.length():b.length();                    }};