Simplify Path

最新推荐文章于 2020-02-07 13:43:23 发布

转载最新推荐文章于 2020-02-07 13:43:23 发布 · 83 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zhutianpeng/p/4263414.html

本文介绍了一个简化Unix风格绝对路径的方法，通过使用栈来解析并去除多余的符号如'.'和'..'，最终返回规范化的路径字符串。

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题目：

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".

代码实现（未优化）：

#include <iostream>
#include <stack>
#include <string>
using namespace std;
class Solution {
public:
    string simplifyPath(string path) {
		return SplitString(path,"/");
    }
	string SplitString(const string& s, const string& c)
	{
		stack<string>st;
		int pos2 = s.find(c);
		int pos1 = 0;
		while(string::npos != pos2)
		{
			string te = s.substr(pos1, pos2-pos1);
			
			if(te.length() > 0){
				
				if(te.compare("..") == 0){
					if(!st.empty())
						st.pop();
				}else if(te.compare(".") == 0){}
				else{
					st.push(te);
				}
			}
			pos1 = pos2 + c.size();
			pos2 = s.find(c, pos1);
		}
		if(pos1 != s.length()){
			string te = s.substr(pos1);
			if(te.compare("..") == 0){
				if(!st.empty())
					st.pop();
			}else if(te.compare(".") == 0){}
			else{
				st.push(te);
			}
		}
		string r = "";
		while(st.empty() == NULL){
			r =  "/" + st.top() + r ;
			st.pop();
		}
		if(r.compare("") == 0){
			r = "/";
		}
		return r;
	}
};
int main(){
	Solution s = Solution();
	string r = s.simplifyPath("///");
	cout << r << endl;
}

转载于:https://www.cnblogs.com/zhutianpeng/p/4263414.html