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: logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"] Output: ["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"]
Constraints:
0 <= logs.length <= 1003 <= logs[i].length <= 100logs[i]is guaranteed to have an identifier, and a word after the identifier.
这道题让给日志排序，每条日志是由空格隔开的一些字符串，第一个字符串是标识符，可能由字母和数字组成，后面的是日志的内容，只有两种形式的，要么都是数字的，要么都是字母的。排序的规则是对于内容是字母的日志，按照字母顺序进行排序，假如内容相同，则按照标识符的字母顺序排。而对于内容的是数字的日志，放到最后面，且其顺序相对于原顺序保持不变。博主感觉这道题似曾相识啊，貌似之前在很多 OA 中见过，最后还是被 LeetCode 收入囊中了。其实这道题就是个比较复杂的排序的问题，两种日志需要分开处理，对于数字日志，不需要排序，但要记录其原始顺序。这里就可以用一个数组专门来保存数字日志，这样最后加到结果 res 后面，就可以保持其原来顺序。关键是要对字母型日志进行排序，同时还要把标识符提取出来，这样在遍历日志的时候，先找到第一空格的位置，这样前面的部分就是标识符了，后面的内容就是日志内容了，此时判断紧跟空格位置的字符，假如是数字的话，说明当前日志是数字型的，加入数组 digitLogs 中，并继续循环。如果不是的话，将两部分分开，存入到一个二维数组 data 中。之后要对 data 数组进行排序，并需要重写排序规则，要根据日志内容排序，若日志内容相等，则根据标识符排序。最后把排序好的日志按顺序合并，存入结果 res 中，最后别忘了把数字型日志也加入 res， 参见代码如下：
class Solution {
public:
vector<string> reorderLogFiles(vector<string>& logs) {
vector<string> res, digitLogs;
vector<vector<string>> data;
for (string log : logs) {
auto pos = log.find(" ");
if (log[pos + 1] >= '0' && log[pos + 1] <= '9') {
digitLogs.push_back(log);
continue;
}
data.push_back({log.substr(0, pos), log.substr(pos + 1)});
}
sort(data.begin(), data.end(), [](vector<string>& a, vector<string>& b) {
return a[1] < b[1] || (a[1] == b[1] && a[0] < b[0]);
});
for (auto &a : data) {
res.push_back(a[0] + " " + a[1]);
}
for (string log : digitLogs) res.push_back(log);
return res;
}
};
参考资料：
https://leetcode.com/problems/reorder-data-in-log-files/
[LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)
来源：https://www.cnblogs.com/grandyang/p/12596394.html