953. Verifying an Alien Dictionary

2019/11/22 Leetcode

953. Verifying an Alien Dictionary

Tags: ‘Hash Table’

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. 1 <= words.length <= 100
  2. 1 <= words[i].length <= 20
  3. order.length == 26
  4. All characters in words[i] and order are english lowercase letters.

Solution

public boolean isAlienSorted(String[] words, String order) {

    // We can also use array for hash, since char set is predefined
    // int[] index = new int[26];
    // for (int i = 0; i < order.length(); ++i)
    //     index[order.charAt(i) - 'a'] = i;

    
    Map<Character, Integer> map = new HashMap<>();
    for (int i = 0; i < order.length(); i++) {
        char c = order.charAt(i);
        map.put(c, i);
    }
    
    for (int i = 0; i < words.length - 1; i++) {
        String before = words[i]; 
        String after = words[i+1];
        
        for (int j = 0; j < before.length(); j++) { 
            // iterate through before
            // if before ends, after doesn't matter
            int index1 = map.get(before.charAt(j));
            if (j >= after.length()) {
                return false;
            }
            int index2 = map.get(after.charAt(j));
            if (index1 < index2) {
                break; // valid. Continue to next word;
            } else if (index1 > index2) {
                return false; // not valid, just return
            } else {
                continue; // equal, continue check next char
            }
            
        }
    }
    return true;
}

Search

    Table of Contents

    文章目录