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19.2.13 [LeetCode 71] Simplify Path

res 根目錄 要求 targe display alt fin per bsp

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"

題意

簡化絕對路徑,最後一個目錄後不能有/

空路徑寫作"/",如果要求訪問根目錄的上一級目錄則自動簡化為根目錄自己,如例2

題解

技術分享圖片
 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

19.2.13 [LeetCode 71] Simplify Path