71 Simplify Path

本文介绍了一个算法问题:如何将Unix风格的文件系统中的绝对路径简化为规范路径。通过解析输入的路径,去除多余的符号如..和..,并确保返回的路径符合规范要求。

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1 题目

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"

2 标准解

class Solution {
public:
    string simplifyPath(string path) {
        char copy[path.size()+1];
        strcpy(copy,path.c_str());
        char *p;
        p = strtok(copy,"/");
        stack<char*> saver;
        while(p != NULL){
            string temp = string(p);
            if(temp == "..") {
                if(!saver.empty()) saver.pop();
            }
            else if(temp!=".") saver.push(p);
            p = strtok(NULL,"/");
        }
        if(saver.empty()) return "/";
        string result;
        while(!saver.empty()){
            result = "/" + string(saver.top()) + result;
            saver.pop();
        }
        return result;
    }
};

 

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