最长非公共字符串 Longest Uncommon Subsequence I

问题：

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.

解决：

【注】理解题意：若两个字符串相等，表示它们彻底相同，没有不一样的子序列，返回-1。若两个字符串有一点不一样，则返回长度更长的那一个。

① 直接查找。

public class Solution {//5ms
    public int findLUSlength(String a, String b) {
        int len1 = a.length();
        int len2 = b.length();
        if(len1 != len2) return Math.max(len1,len2);
        else{
            for(int i = 0;i < len1;i ++){
                if(a.charAt(i) != b.charAt(i)) return len1;
            }
        }
        return -1;
    }
}

② 进化版

public class Solution {//3ms
    public int findLUSlength(String a, String b) {
        if (a == null) {
            if (b == null) {
                return -1;
            }
            return b.length();
        }
        if (b == null) {
            return a.length();
        }      
        return (a.equals(b) ? -1 : Math.max(a.length(), b.length()));
    }
}

③ 最直接的作法

public class Solution { //5ms
    public int findLUSlength(String a, String b) {
        if(! a.equals(b)) return Math.max(a.length(),b.length());
        return -1;
    }
}

public class Solution { //3ms     public int findLUSlength(String a, String b) {         if (a.equals(b))             return -1;         return Math.max(a.length(), b.length());     } }