521. Longest Uncommon Subsequence I

Description

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:

1.Both strings’ lengths will not exceed 100.

2.Only letters from a ~ z will appear in input strings.

这道题要求返回两个字符串的最长非平凡子序列，在题目的描述中，对最长非平凡子序列给出了如下的定义：

最长非平凡子序列是指：某个字符串的子字符串，不是另一个字符串的子串。

具体的例子见题干中给出的Example。

这种题很适合采用“测试驱动开发”的原则，给出下列测试用例：

1.“aba”,”cdc”,由于两者之间互不为子串，所以返回两者之中的最长字符串的长度即可

2.“aba”,”babac”,由于aba是后者的子串，但后者却不是前者的子串，所以返回较长的即可

3.两者中有一个为空串，由于题目说明空串是任何字符串的子串，所以依然返回较长的即可

4.两字符串相等，如“aaa”,”aaa”,找不到某一个子串同时不为另一个字符串的子串，所以返回-1

总结一下上述情况，其实就是两个字符串不相等时返回较长的，相等时返回-1，代码如下:

public class Solution {
    public int findLUSlength(String a, String b) {
        if(!b.contains(a)&&!a.contains(b))
        	return Math.max(a.length(), b.length());
        if(a.contains(b)&&b.contains(a))
        	return -1;
        if(a.contains(b)||b.contains(a))
        	return Math.max(a.length(), b.length());
        if(("".equals(a)&&!"".equals(b))||(!"".equals(a)&&"".equals(b)))
        	return Math.max(a.length(), b.length());
        else
        	return 0;
    }
}

one-line版本的如下：

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