共计 1872 个字符，预计需要花费 5 分钟才能阅读完成。
题目
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).
Note:
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
All characters in words[i] and order are english lowercase letters.
题目地址
讲解
这道题讲道理，我觉得有点难。可能递归的解法比较简单吧。刚开始我思路一直是错的，正确的思路应该是先深入比较两个字符串，没有问题再比较下一对字符串（这是一种深度优先，我刚开始一直是广度优先的思路）。主要的麻烦来自于对数组越界的判断。
java 代码
class Solution {
public boolean isAlienSorted(String[] words, String order) {
if(words.length==1){
return true;
}
Map<Character, Integer> map = new HashMap<>();
for(int i=0;i<order.length();i++){
map.put(order.charAt(i), i);
}
for(int i=1;i<words.length;i++){
int index = 0;
boolean indexOut = false;
if(index>words[i-1].length()-1){
continue;
}
if(index>words[i].length()-1){
return false;
}
while(map.get(words[i].charAt(index))==map.get(words[i-1].charAt(index))){
index++;
System.out.println(index);
if(index>words[i-1].length()-1){
indexOut=true;
break;
}
if(index>words[i].length()-1){
return false;
}
}
if(indexOut){
continue;
}
if(map.get(words[i].charAt(index))<map.get(words[i-1].charAt(index))){
return false;
}
}
return true;
}
}