953. Verifying an Alien Dictionary*
https://leetcode.com/problems/verifying-an-alien-dictionary/
题目描述
In an alien language, surprisingly they also use english lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.
Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographicaly in this alien language.
Example 1:
Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
Constraints:
-
1 <= words.length <= 100 -
1 <= words[i].length <= 20 -
order.length == 26 - All characters in
words[i] and order are English lowercase letters.
C++ 实现 1
这题关键在于如何判断两个字符串 a 和 b 是 lexicographicaly 有序的:
- 如果
a[i] > b[i], 说明不是字典序 - 如果存在
a[i] < b[i] , 那么就是字典序 - 存在
xxxy 和xxx 的情况, 前半部分字符相等, 但是a 比b 长, 仍然不是字典序.
class Solution {
private:
unordered_map<char, int> record;
bool isSorted(const string &before, const string &after) {
int i = 0;
while (i < before.size() && i < after.size()) {
if (record[before[i]] < record[after[i]]) return true;
else if (record[before[i]] > record[after[i]]) return false;
else ++ i;
}
// 循环跳出说明两个字符串前面部分的字符完全相等, 如果此时 before 更长的话,
// 仍然要返回 false
if (before.size() > after.size()) return false;
return true;
}
public:
bool isAlienSorted(vector<string>& words, string order) {
if (words.size() <= 1) return true;
for (int i = 0; i < order.size(); ++ i) record[order[i]] = i;
for (int i = 1; i < words.size(); ++ i) {
if (!isSorted(words[i - 1], words[i])) return false;
}
return true;
}
};
