- Question
- Ideas
- 1、Answer( Java )
- Code
937. 重新排列日志文件
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reorder-data-in-log-files/
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Ideas 1、Answer( Java )
解法思路:简单自定义排序
Code/**
* @author Listen 1024
* @description 937. 重新排列日志文件 (简单模拟排序题)
* 时间复杂度 O(nlogn)
* 空间复杂度 O(n)
* @date 2022-05-03 12:04
*/
class Solution {
public String[] reorderLogFiles(String[] logs) {
String[] res = new String[logs.length];
ArrayList<String> listNumDig = new ArrayList<>();
ArrayList<String> listStrDig = new ArrayList<>();
for (String log : logs) {
String[] strings = log.split("\s+");
int i = strings[1].charAt(0);
if (i >= 48 && i <= 57) {//judge the num from [0,9] of the ASCII
listNumDig.add(log);
} else {
listStrDig.add(log);
}
}
listStrDig.sort(new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
int index1 = o1.indexOf(" ");
int index2 = o2.indexOf(" ");
String temp1 = o1.substring(index1, o1.length());
String temp2 = o2.substring(index2, o2.length());
if (!temp1.equals(temp2)) {
return temp1.compareTo(temp2);
} else {
return o1.substring(0, index1).compareTo(o2.substring(0, index2));
}
}
});
int index = 0;
for (int i = 0; i < listStrDig.size(); i++) {
res[i] = listStrDig.get(i);
index++;
}
for (int i = 0; i < listNumDig.size(); i++) {
res[index] = listNumDig.get(i);
index++;
}
return res;
}
}
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