zoukankan      html  css  js  c++  java
  • 937. Reorder Log Files

    You have an array of logs.  Each log is a space delimited string of words.

    For each log, the first word in each log is an alphanumeric identifier.  Then, either:

    • Each word after the identifier will consist only of lowercase letters, or;
    • Each word after the identifier will consist only of digits.

    We will call these two varieties of logs letter-logs and digit-logs.  It is guaranteed that each log has at least one word after its identifier.

    Reorder the logs so that all of the letter-logs come before any digit-log.  The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties.  The digit-logs should be put in their original order.

    Return the final order of the logs.

    Example 1:

    Input: ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
    Output: ["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]

    Note:

    1. 0 <= logs.length <= 100
    2. 3 <= logs[i].length <= 100
    3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

    Approach #1: Sort. [Java]

    class Solution {
        public String[] reorderLogFiles(String[] logs) {
            Comparator<String> myComp = new Comparator<String>() {
                @Override
                public int compare(String s1, String s2) {
                    int s1si = s1.indexOf(' ');
                    int s2si = s2.indexOf(' ');
                    char s1fc = s1.charAt(s1si+1);
                    char s2fc = s2.charAt(s2si+1);
                    if (s1fc <= '9') {
                        if (s2fc <= '9') return 0;
                        else return 1;
                    }
                    if (s2fc <= '9') return -1;
                    int preCompute = s1.substring(s1si+1).compareTo(s2.substring(s2si+1));
                    if (preCompute == 0) 
                        return s1.substring(0, s1si).compareTo(s2.substring(0, s2si));
                    return preCompute;
                }
            };
            
            Arrays.sort(logs, myComp);
            return logs;
        }
    }
    

      

    Analysis:

    This solution takes advantage of what we're guaranteed in the problem:

    1. guaranteed to have a word following an identifier (allows me to use indexOf(' ‘) freely.).

    2.letter logs need to be ordered lexicographically, so we can use built in compare function when we know we have two.

    3. number logs need to be sorted naturally, so we just say they are all "equal" to each other and trust java's built in sort feature to be stable.

    4. number logs need to be after letter logs, so once we find out they're differenct, we return one of the other and short-circuit.

    Reference:

    https://leetcode.com/problems/reorder-log-files/discuss/193872/Java-Nothing-Fancy-15-lines-5ms-all-clear.

    http://www.cnblogs.com/skywang12345/p/3324788.html

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    xss框架(一)之浏览器通信
    Joomla未授权创建特权用户漏洞和getshell脚本解析
    从零开始写网站登录爆破(一)
    CSRF学习整理
    vue中vue2-google-maps使用谷歌地图的基础操作
    vue中百度地图API的调用
    60秒定时减少
    git操作指令,以及常规git代码操作
    taro taroUi的H5打包后路径/修改为./
    vue enter事件无效,加入native
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10799385.html
Copyright © 2011-2022 走看看