题目描述
Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1.
An uncommon subsequence between an array of strings is a string that is a subsequence of one string but not the others.
A subsequence of a string s is a string that can be obtained after deleting any number of characters from s.
For example, “abc” is a subsequence of “aebdc” because you can delete the underlined characters in “aebdc” to get “abc”. Other subsequences of “aebdc” include “aebdc”, “aeb”, and “” (empty string).
Example 1:
Input: strs = [“aba”,”cdc”,”eae”] Output: 3
Example 2:
Input: strs = [“aaa”,”aaa”,”aa”] Output: -1
Constraints:
- 1 <= strs.length <= 50
- 1 <= strs[i].length <= 10
- strs[i] consists of lowercase English letters.
解题思路
C++代码
class Solution {
public:
int findLUSlength(vector<string>& strs) {
int res = -1;
for(int i = 0; i < strs.size(); i++){
int j = 0;
for(j = 0; j < strs.size(); j++){
if(i == j){
continue;
}
if(check(strs[i], strs[j])){
break;
}
}
if(j == strs.size()){
res = max(res, (int)strs[i].length());
}
}
return res;
}
bool check(string a, string b){
int i = 0;
for(auto& s : b){
if(s == a[i]){
i++;
}
if(i == a.size()){
break;
}
}
return i == a.size();
}
};
