<leetcode> 521. Longest Uncommon Subsequence I-优快云博客

本文探讨了如何寻找两个字符串中的最长非公共子序列，并给出了一种简单有效的解决方案。通过对题目的正确理解，我们发现只需比较两个字符串是否相同，不同则返回较长字符串的长度。

521. Longest Uncommon Subsequence I

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.

一开始的时候，用google translate，翻译的不是很清楚，导致我的理解有问题：一开始我还以为是在A,B字符串中找最长非子串，

也就是A中在B中不是子串的最长子串，和B中在A中不是子串的最长子串,取其中的最长的长度。。

后来发现这样好麻烦，就百度了一下，发现自己的理解有问题。

两个字符串中最长的字符串是否为另一字符串的子字符串，如果是，则返回-1，否则返回最长字符串长度。

就是只需要判断两字符串是否相等，如果相等就返回-1，不相等就返回较长的字符串长度！

怪不得，这题目68赞，984踩。。。。

class Solution {
    public int findLUSlength(String a, String b) {
        int n = a.length();
        int m = b.length();
        if (a.equals(b)) return -1;
        else return n>m ? n : m;
    }
}