Reorder Log Files
String
Easy
You have an array oflogs
. Each log is a space delimited string of words.
For each log, the first word in each log is an alphanumericidentifier. 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 logsletter-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.
Analysis
Write a custom sort (compare function)
class Solution {
public String[] reorderLogFiles(String[] logs) {
Comparator<String> cmp = new Comparator<String>() {
@Override
public int compare(String log1, String log2) {
int i1 = log1.indexOf(" ");
int i2 = log2.indexOf(" ");
char ch1 = log1.charAt(i1 + 1);
char ch2 = log2.charAt(i2 + 1);
if (Character.isDigit(ch1)) {
if (Character.isDigit(ch2)) {
return 0;
} else {
return 1;
}
}
if (Character.isDigit(ch2)) {
return -1;
}
int comp = log1.substring(i1 + 1).compareTo(log2.substring(i2 + 1));
if (comp == 0) {
return log1.substring(0, i1).compareTo(log2.substring(0, i2));
}
return comp;
}
};
Arrays.sort(logs, cmp);
return logs;
}
}
Last updated
Was this helpful?