Simplify Path

最新推荐文章于 2020-02-07 13:43:23 发布

zephyr_be_brave

最新推荐文章于 2020-02-07 13:43:23 发布

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分类专栏： leetcode习题分析

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本文介绍了一种使用双端队列(deque)简化Unix风格文件路径的方法。通过解析输入路径中的符号如'..'和'.'，去除多余的斜杠，最终返回规范化的路径字符串。

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

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

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".

方法一：我用deque实现的，当然也可以用栈

class Solution {
public:
   
    string simplifyPath(string path) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        deque<string> queueS;
	int pos = path.find("/");
        if(pos == string::npos) return path;
		queueS.push_back(path.substr(pos,1));
        int old_pos = pos+1;
        
        
        while((pos = path.find("/",old_pos))!=string::npos){
            
            string temp = path.substr(old_pos,pos-old_pos);
            if(temp == "..") {
                if(queueS.size()>=2)
		{
			queueS.pop_back();
			queueS.pop_back();
		}       
            }else if(temp == ".") {}
            else{
                if(temp != "")
                    queueS.push_back(temp);
            }

	    if(queueS.back() != "/") {
		string s = path.substr(pos,1);
		queueS.push_back(s);
	    }
            old_pos = pos+1;
        }
    
    
        string temp = path.substr(old_pos);
        if(temp == "..") {
            if(queueS.size()>=2)
	    {
		queueS.pop_back();
		queueS.pop_back();
	    }
        }else if(temp == ".") {}
        else{
            if(temp != "")
                queueS.push_back(temp);
        }
        
	if(queueS.size()>1 && queueS.back()=="/") queueS.pop_back();
        
        string ret;
        while(!queueS.empty())
        {
            ret.append(queueS.front());
            queueS.pop_front();
        }
        
        return ret;
    }
};

36 milli secs.