一、題目描述

　　

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.

　　解釋：

　　給定一個字符串的數組，每個字符串中間用空格分割，第一個元素表示索引，后面的為內容；內容分為兩種，一種是純字符串，一種是純數組

　　要求進行排序，排序規則是，純字符串內容的排在純數字的前面；對于純字符串的元素，比較索引大小排序；對于純數字的其相對位置不變

　　

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; }

　

三、學習到的方法

　　sort使用的快速排序；stable_sort使用的是merge排序，穩定的，意思是相等的元素前后的位置會保持

　　

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　　另外，string的兩個方法

　　1、substr(index) ,從0到index的子串

　　2、find('') ,表示尋找某個字符所在的位置

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轉載于:https://www.cnblogs.com/doudouyoutang/p/10189345.html

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