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:

1. Both strings' lengths will not exceed 100.
2. Only letters from a ~ z will appear in input strings.

A:

好像这个是我理解错了题意了。--------理解成了  不属于 第二个的任意一个-----

class Solution {
public:
    int findLUSlength(string a, string b) {
        unordered_map<char, int> Ma;
        for(auto ch : a)
            Ma[ch] = Ma.find(ch)==Ma.end()? 1 : Ma.find(ch)->second +1;
                
        unordered_map<char, int> Mb;
        for(auto ch : b)
            Mb[ch] = Mb.find(ch)==Mb.end()? 1 : Mb.find(ch)->second +1;
        
        int resA =0;
        for(auto it : Ma)
        {
            if(Mb.find(it.first) == Mb.end())
            {
                resA += it.second;
            }
        }
        int resB =0;
        for(auto it : Mb)
        {
            if(Ma.find(it.first) == Ma.end())
            {
                resB += it.second;
            }
        }
        int res = max(resA, resB);
        return res == 0?-1:res;
    }
};

-------------------------其实是，只要不是第二个的subsequence就可以，而不是不包含---------

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