953. Verifying an Alien Dictionary

题目

In an alien language, surprisingly they also use english lowercase letters, but possibly in a different order. The orderof 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.

题目大意

某种外星语也使用英文小写字母，但可能顺序 order 不同。字母表的顺序（order）是一些小写字母的排列。给定一组用外星语书写的单词 words，以及其字母表的顺序 order，只有当给定的单词在这种外星语中按字典序排列时，返回 true；否则，返回 false。

解题思路

这一题是简单题。给出一个字符串数组，判断把字符串数组里面字符串是否是按照 order 的排序排列的。order 是给出个一个字符串排序。这道题的解法是把 26 个字母的顺序先存在 map 中，然后依次遍历判断字符串数组里面字符串的大小。

代码


package leetcode

func isAlienSorted(words []string, order string) bool {
	if len(words) < 2 {
		return true
	}
	hash := make(map[byte]int)
	for i := 0; i < len(order); i++ {
		hash[order[i]] = i
	}
	for i := 0; i < len(words)-1; i++ {
		pointer, word, wordplus := 0, words[i], words[i+1]
		for pointer < len(word) && pointer < len(wordplus) {
			if hash[word[pointer]] > hash[wordplus[pointer]] {
				return false
			}
			if hash[word[pointer]] < hash[wordplus[pointer]] {
				break
			} else {
				pointer = pointer + 1
			}
		}
		if pointer < len(word) && pointer >= len(wordplus) {
			return false
		}
	}
	return true
}