[LeetCode] Simplify Path，文件路径简化，用栈来做

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".

如果仅仅从不断replace输入路径的角度出发，会非常复杂。

如果用一个栈来一次存储各级路径的directory名字，然后重组，会简便一些，这也是文件路径简化类题目的常用思路。

为了末尾可以顺序遍历栈重组path，我们不用传统的stack lib，而用vector来实现栈，这样可以方便顺序遍历。

代码：

class Solution {

public:

    string simplifyPath(string path) {

        if(path.length() == 0) return "";

        vector<string> v;

        int p = 0, start = 0;

        while(p < path.length()){

            if(path[p] == '/') ++p;

            else{

                start = p;

                while(p < path.length() && path[p] != '/') ++p;

                string temp = path.substr(start, p-start);

                if(temp == ".."){ 

                    if(!v.empty()) v.pop_back(); //遇到".."就出栈

                    else{

                        if(path[0] != '/') v.push_back(temp); //如果没东西可以出，就要看path是否以根目录开头了，不是以根目录开头的话，".."要进栈的

                    }

                }else if(temp == "."){} //遇到"."就跳过

                else if(temp.length() > 0){

                    v.push_back(temp);

                }

            }

        }

        string res = (path[0] == '/' ? "/" : ""); //重组path

        for(vector<string>::iterator i = v.begin(); i < v.end(); ++i){

            res.append(*i);

            if(i < v.end()-1) res.append("/");

        }

        return res;

    }

};