LeetCode - Verifying an Alien Dictionary [난이도:Easy]

문제 설명

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.

입출력 예시

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).

제약사항

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

Solution

let isAlienSorted = function(words, order) {
    let map = new Map();

    // 알파벳을 key, 순서를 value로 하여 순서 비교에 사용
    for(let i=0; i<order.length; i++){
        map.set(order[i],i);
    }

    // 앞에서 부터 두단어씩 비교
    for(let j=0; j<words.length-1; j++){
        let preWord=words[j];
        let nextWord=words[j+1];
        
        let lastIdxofnw=nextWord.length-1;

        for(k=0; k<preWord.length; k++){
            
            // 앞자리까지 순서가 같은데 앞단어에는 추가문자가 있을경우는 잘못된 정렬이므로 return false
            if(k>lastIdxofnw)
            {
                return false;
            }

            // 순서를 수치화한 값을 비교하여, 
            if(map.get(preWord[k])<map.get(nextWord[k])){ //앞단어의 문자 순서가 우선시되면 다음 단어 간 비교로 넘어감
               break;
            }
            else if(map.get(preWord[k])>map.get(nextWord[k])){ //뒤 단어의 문자의 순서가 우선시되는 거면 return false;
                return false;
            }
            //같은번째 자리에 같은 문자면 이후도 봐야하므로 for문 계속
        }
    }

    //검증을 다 통과했으면 return true;

    return true;
};

^~^