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LeetCode:Simplify Path

题目链接

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".

分析:需要注意的是/…可以表示名字为…的路径,路径的最后可能没有/。可以利用栈,碰到正常路径压入栈中,碰到/.不作任何操作,碰到/..删除栈顶元素。下面代码中用数组来模拟栈                                                                                  本文地址

class Solution {
public:
    string simplifyPath(string path) {
        int len = path.size();
        vector<string> vec;
        int i = 0, index = 0;
        while(i < len)
        {
            int j = path.find('//', i + 1);
            string tmp;
            if(j != string::npos)
                tmp = path.substr(i, j - i);
            else {tmp = path.substr(i, len); j = len;}
        
            if(tmp == "/");
            else if(tmp == "/.");
            else if(tmp == "/..")
                {if(!vec.empty())vec.pop_back();}
            else 
                vec.push_back(tmp);
            i = j;
        }
        if(vec.empty())return "/";
        else 
        {
            string res;
            for(int i = 0; i < vec.size(); i++)
                res += vec[i];
            return res;
        }
    }
}; 

 

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posted @ 2013-12-09 14:14  tenos  阅读(1241)  评论(1编辑  收藏  举报

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