标签：nal after case 字符串 输入 static targe 包括 eric
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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:
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 <= 1003 <= logs[i].length <= 100logs[i] is guaranteed to have an identifier, and a word after the identifier.
大意：给你一些日志，包括英文日志和数字日志，每个日志又包括日志头和日志内容，日志头是第一个单词，日志内容全数字的是数字日志，全英文的是英文日志。要求排序后输出。
排序规则：对英文的日志，去掉日志头，按日志内容字典序排序。对数字的日志，按输入顺序。
总体上，英文日志在前，数字日志在后。
思路：对每个字符串的最后一位进行判断之后，分类到两个向量里，对英语日志用stringstream进行分割，然后sort排序。
代码：
class Solution { public: static bool cmp(string s1,string s2){ string news1="",news2=""; stringstream ss1(s1); string s; int k = 0; while(ss1>>s){ if(k > 0){ news1 = news1 + s +" "; } k++; } k = 0; stringstream ss2(s2); while(ss2>>s){ if(k > 0){ news2 = news2 + s +" "; } k++; } if(news1<news2) return true; return false; //return news1 < news2; } vector<string> reorderLogFiles(vector<string>& logs) { vector<string>p1,p2,ans; for(int i = 0 ; i < logs.size() ; i++){ string s = logs[i]; if(s[s.size()-1]<=‘9‘ && s[s.size()-1] >= ‘0‘){ p2.push_back(s); } else{ p1.push_back(s); } } sort(p1.begin(),p1.end(),cmp); for(int i = 0 ; i < p1.size() ; i ++){ ans.push_back(p1[i]); } for(int i = 0 ; i < p2.size() ; i ++){ ans.push_back(p2[i]); } return ans; } };
leecode 937 Reorder Log Files （模拟）
标签：nal after case 字符串 输入 static targe 包括 eric
原文地址：https://www.cnblogs.com/Esquecer/p/10325217.html
