leetcode讲解--937. Reorder Log Files

题目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:0 <= logs.length <= 1003 <= logs[i].length <= 100logs[i] is guaranteed to have an identifier, and a word after the identifier.题目地址讲解这道题我看过后立马联想到了另一个题目：953. Verifying an Alien Dictionary。那个题目是判断是否有序，而现在这个题目是要排序。感觉这一题更难。实际做下来确实还是比较难的，主要是字符串的比较，要自己写比较规则。然后在调用系统的sort函数。如果要自己写快排那又要多花不少时间了。有个小窍门，题目要求数字的logs保持原样，而且要放在数组尾部，所以我这里是从后往前遍历数组。Java代码class Solution { public String[] reorderLogFiles(String[] logs) { String[] result = new String[logs.length]; int indexBegin = 0; int indexEnd = logs.length-1; for(int i=logs.length-1;i>=0;i–){ int firstSpaceIndex = logs[i].indexOf(’ ‘); if(logs[i].charAt(firstSpaceIndex+1)>=‘0’ && logs[i].charAt(firstSpaceIndex+1)<=‘9’){ result[indexEnd–] = logs[i]; }else{ result[indexBegin++] = logs[i]; } } // for(int i=0;i<logs.length;i++){ // System.out.println(result[i]); // } StringComparator comparator = new StringComparator(); Arrays.sort(result, 0, indexBegin, comparator); return result; } class StringComparator implements Comparator<String>{ @Override public int compare(String o1, String o2) { int o1IndexOfFirstSpace = o1.indexOf(’ ‘); o1 = o1.substring(o1IndexOfFirstSpace); int o2IndexOfFirstSpace = o2.indexOf(’ ‘); o2 = o2.substring(o2IndexOfFirstSpace); int minLength = o1.length(); boolean o1ShortThano2 = true; if(o1.length()>o2.length()){ minLength = o2.length(); o1ShortThano2 = false; } boolean o1LittleThano2 = true; boolean o1Equalso2 = true; for(int i=0;i<minLength;i++){ if(o1.charAt(i)<o2.charAt(i)) { o1Equalso2 = false; break; }else if(o1.charAt(i)>o2.charAt(i)){ o1LittleThano2 = false; o1Equalso2 = false; break; } } if(o1Equalso2){ if(o1ShortThano2){ return -1; }else{ return 1; } }else{ if(o1LittleThano2){ return -1; }else{ return 1; } } } }}