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.

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

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());
}
}

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