[LeetCode] 937. Reorder Log Files

nemo 发布于2019-08-16 14:25 / 1658人阅读

Problem

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.

Solution

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        Comparator comparator = new Comparator() {
            @Override
            public int compare(String s1, String s2) {
                int i1 = s1.indexOf(" ");
                int i2 = s2.indexOf(" ");
                char c1 = s1.charAt(i1+1);
                char c2 = s2.charAt(i2+1);
                if (c1 <= "9") {
                    return c2 <= "9" ? 0 : 1;
                } else if (c2 <= "9") {
                    return -1;
                } else {
                    //all letters
                    int res = s1.substring(i1+1).compareTo(s2.substring(i2+1));
                    if (res == 0) return s1.substring(0,i1).compareTo(s2.substring(0,i2));
                    else return res;
                }
            }
        };
        
        Arrays.sort(logs, comparator);
        return logs;
    }
}

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