leetcode-953-easy

Verifying an Alien Dictionary

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 lexicographically 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.

思路一：先把字符顺序弄成 26 个数组，然后用这个顺序对单词进行排序，最后查看排序后的数组是否和原数组匹配。看了一下，运行效率比较低，官方解答比较好，不用全部进行比较，只要对数组进行遍历比较

{
        char[] chars = order.toCharArray();
        int[] sortedChars = new int[26];

        for (int i = 0; i < chars.length; i++) {
            sortedChars[chars[i] - 'a'] = i;
        }

        String[] clone = words.clone();

        Arrays.sort(words, (o1, o2) -> {
            char[] chars1 = o1.toCharArray();
            char[] chars2 = o2.toCharArray();

            int min = Math.min(chars1.length, chars2.length);
            for (int i = 0; i < min; i++) {
                if (sortedChars[chars1[i] - 'a'] == sortedChars[chars2[i] - 'a']) {
                    continue;
                }

                return sortedChars[chars1[i] - 'a'] < sortedChars[chars2[i] - 'a'] ? -1 : 1;
            }

            return chars1.length < chars2.length ? -1 : 1;
        });

        return Arrays.equals(clone, words);
    }