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.
digit-log = identifier + 小寫字母組成
letter-logs = identifier + 數字組成
至少有一個字
請排序 letter-logs + digit-log
letter-logs 不看第一個 word 其他按照 a-z 排，digit-logs 看本來怎麼排就依照原本順序
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 <= 100
3 <= logs[i].length <= 100
logs[i] is guaranteed to have an identifier,
and a word after the identifier.
怎麼解
先判斷是 digit-log 還是 letter-logs
letter-logs就去抓第 2 個 word 排序
digit-log 就是 queue 抓到存，最後再 concat 到 letter-logs
var reorderLogFiles = function(logs) {
let digitArr = [];
let letterArr = []
logs.forEach(log => {
let x = 0
while(log[x]!==' '){
x++;
}
// console.log(Number(log[x + 1]))
if(Number(log[x + 1]) == log[x + 1]){
digitArr.push(log);
}else{
letterArr.push(log)
}
})
letterArr.sort( (a, b) => {
let strA = a.substr(a.indexOf(' ') + 1);
let strB = b.substr(b.indexOf(' ') + 1);
if(strA === strB) return a.localeCompare(b);
return strA.localeCompare(strB);
})
return [...letterArr, ...digitArr]
};