zoukankan      html  css  js  c++  java
  • 0953. Verifying an Alien Dictionary (E)

    Verifying an Alien Dictionary (E)

    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).
    

    Constraints:

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

    题意

    给定一个自定义的字典序,判断一个字符串数组是否递增。

    思路

    Hash的应用。


    代码实现

    Java

    class Solution {
        public boolean isAlienSorted(String[] words, String order) {
            int[] priority = new int[26];
    
            for (int i = 0; i < 26; i++) {
                priority[order.charAt(i) - 'a'] = i;
            }
    
            for (int i = 0; i < words.length - 1; i++) {
                if (compare(words[i], words[i + 1], priority) == 1) return false;
            }
    
            return true;
        }
    
        private int compare(String a, String b, int[] priority) {
            for (int i = 0; i < Math.min(a.length(), b.length()); i++) {
                char ca = a.charAt(i), cb = b.charAt(i);
                if (priority[ca - 'a'] > priority[cb - 'a']) {
                    return 1;
                } else if (priority[ca - 'a'] < priority[cb - 'a']) {
                    return -1;
                }
            }
            return a.length() == b.length() ? 0 : a.length() > b.length() ? 1 : -1;
        }
    }
    
  • 相关阅读:
    单片机编程积累算法
    关于GSM基站定位
    GSM模块fibocom G510使用记录
    指爱 打字比赛记录
    硬件和软件工程师
    GPS模块启动模式说明
    阻容降压电路分析
    饮水机电路-工作剖析
    跑步,去
    day01 IT知识架构,操作系统简介
  • 原文地址:https://www.cnblogs.com/mapoos/p/14637416.html
Copyright © 2011-2022 走看看