共计 2265 个字符，预计需要花费 6 分钟才能阅读完成。
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 <= 100
3 <= logs[i].length <= 100
logs[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;
}
}
}
}
}
