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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();
				    }
			}
			else
				name = 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Accepted
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();
				    }
			}
			else
				name = 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;
    }
};

leetcode_71_Simplify Path

标签：c++   leetcode   stack   

原文地址：http://blog.csdn.net/keyyuanxin/article/details/43699497