leetcode_71_Simplify Path

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Simplify Path
Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"
click to show corner cases.

Corner Cases:
Did you consider the case where path = "/../"?
In this case, you should return "/".
Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
In this case, you should ignore redundant slashes and return "/home/foo".

思路解析：
面对字符串题目一定要保持清醒，先分析好题目之后再开始码代码。
题目的要求是输出Unix下的最简路径，Unix文件的根目录为"/"，"."表示当前目录，".."表示上级目录。
例如：
输入1：
/../a/b/c/./.. 
输出1：
/a/b
模拟整个过程：
1. "/" 根目录
2. ".." 跳转上级目录，上级目录为空，所以依旧处于 "/"
3. "a" 进入子目录a，目前处于 "/a"
4. "b" 进入子目录b，目前处于 "/a/b"
5. "c" 进入子目录c，目前处于 "/a/b/c"
6. "." 当前目录，不操作，仍处于 "/a/b/c"
7. ".." 返回上级目录，最终为 "/a/b"

实现方法：用一个堆栈来模拟路径的行为，遇到"."不操作，遇到".."退栈，其他情况都压入堆栈。

//vs2012测试代码#include<iostream>#include<stack>#include<string>using namespace std;class Solution {private:    void pathToDirectories( stack<string> &directories , string &path){string name;name.clear();path = path + '/';for(int i=0; i<path.length(); i++){if( path[i]=='/' ){if( !name.empty() )    if ( name[0]=='.' && name.length()==1 )        name.clear();else if( name[0]=='.' && name.length()==2 && name[1]=='.' ){if( !directories.empty() )directories.pop();name.clear();}else {directories.push(name);name.clear();    }}elsename = name + path[i];}}void directoriesToPath( string &ans,stack<string> &directories){ans.clear();while( !directories.empty() ){ans = directories.top() + '/'+ ans ;directories.pop();}if( !ans.empty() )ans = ans.substr( 0,ans.length()-1 );//substr 方法:返回一个从指定位置开始，并具有指定长度的子字符串。ans = '/' + ans;}public:    string simplifyPath(string path) {stack<string> directories;pathToDirectories( directories,path );string ans;directoriesToPath ( ans,directories );return ans;    }};int main(){string path;getline( cin , path);  //或者cin>>s;Solution lin;cout<<lin.simplifyPath(path)<<endl;return 0;}

//方法一：自测Acceptedclass Solution {private:    void pathToDirectories( stack<string> &directories , string &path){string name;name.clear();path = path + '/';for(int i=0; i<path.length(); i++){if( path[i]=='/' ){if( !name.empty() )    if ( name[0]=='.' && name.length()==1 )        name.clear();else if( name[0]=='.' && name.length()==2 && name[1]=='.' ){if( !directories.empty() )directories.pop();name.clear();}else {directories.push(name);name.clear();    }}elsename = name + path[i];}}void directoriesToPath( string &ans,stack<string> &directories){ans.clear();while( !directories.empty() ){ans = directories.top() + '/'+ ans ;directories.pop();}if( !ans.empty() )ans = ans.substr( 0,ans.length()-1 ); ////substr 方法:返回一个从指定位置开始，并具有指定长度的子字符串。ans = '/' + ans;}public:    string simplifyPath(string path) {stack<string> directories;pathToDirectories( directories,path );string ans;directoriesToPath ( ans,directories );return ans;    }};