19.2.13 [LeetCode 71] Simplify Path
作者:互联网
Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.
In a UNIX-style file system, a period .
refers to the current directory. Furthermore, a double period ..
moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix
Note that the returned canonical path must always begin with a slash /
, and there must be only a single slash /
between two directory names. The last directory name (if it exists) must not end with a trailing /
. Also, the canonical path must be the shortest string representing the absolute path.
Example 1:
Input: "/home/" Output: "/home" Explanation: Note that there is no trailing slash after the last directory name.
Example 2:
Input: "/../" Output: "/" Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:
Input: "/home//foo/" Output: "/home/foo" Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Example 4:
Input: "/a/./b/../../c/" Output: "/c"
Example 5:
Input: "/a/../../b/../c//.//" Output: "/c"
Example 6:
Input: "/a//b////c/d//././/.." Output: "/a/b/c"
题意
简化绝对路径,最后一个目录后不能有/
空路径写作"/",如果要求访问根目录的上一级目录则自动简化为根目录自己,如例2
题解
1 class Solution { 2 public: 3 string simplifyPath(string path) { 4 string ans = ""; 5 int p = 0, l = path.length(); 6 while (p < l) { 7 if (path[p] != '/') { 8 string menu = ""; 9 while (p < l&&path[p] != '/') 10 menu += path[p++]; 11 if (menu == ".") 12 continue; 13 else if (menu == "..") { 14 int idx = ans.rfind('/'); 15 if (idx == string::npos)continue; 16 ans.erase(idx); 17 } 18 else 19 ans += "/" + menu; 20 } 21 p++; 22 } 23 if (ans.empty())return "/"; 24 return ans; 25 } 26 };View Code
标签:13,string,..,Input,Simplify,ans,path,19.2,Example 来源: https://www.cnblogs.com/yalphait/p/10371323.html