题目描述

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.

解法一

思路

这题算是专门为愚人节准备的了,如果不仔细理解题意,很容易被绕进去。

题目要求两个字符串的“最长非公共子序列”的长度,怎么求最长?如果两个字符串不相等,那其实任意一个字符串本身都是非公共子序列;所以结果也出来了,如果字符串不想等,那么“最长非公共子序列”就是较长的那个字符串了。如果字符串相等,可想而知不可能找到“最长非公共子序列”。

Python

class Solution(object):
    def findLUSlength(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: int
        """
        return -1 if a == b else max(len(a), len(b))

Java

public class Solution {
    public int findLUSlength(String a, String b) {
        return a.equals(b) ? -1 : Math.max(a.length(), b.length());
    }
}

C++

class Solution {
public:
    int findLUSlength(string a, string b) {
        return a == b ? -1 : max(a.length(), b.length());
    }
};