Leetcode-937-Reorder Log Files-(Easy)

一、题目描述

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.

　　解释：

　　给定一个字符串的数组，每个字符串中间用空格分割，第一个元素表示索引，后面的为内容；内容分为两种，一种是纯字符串，一种是纯数组

　　要求进行排序，排序规则是，纯字符串内容的排在纯数字的前面；对于纯字符串的元素，比较索引大小排序；对于纯数字的其相对位置不变

二、解答：

class Solution {
public:
static bool cmp(const string &A, const string& B){
    string subA = A.substr(A.find(' ') + 1);
    string subB = B.substr(B.find(' ') + 1);
    if(isdigit(subA[0]))
        return false;
    else if(isdigit(subB[0]))
        return true;
    return subA.compare(subB) < 0;
}
 
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
         
        stable_sort(logs.begin(), logs.end(), cmp);
 
        return logs;
    }
};
 
 
int main(int argc, char **argv)
{
    Solution *s = new Solution();
    cout<<""<<endl;
 
    return 0;
}