Leetcode NO.71 Simplify Path

本题题目于要求如下：

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".

本题的难度并不高，对我来说，难度主要集中在unix-style command的含义。。后来搞明白了：

1. "."的意思是忽略，不做任何处理。
2. ".."的意思是返回上一级，具体在本题的stack的操作是，pop栈顶的元素，由于栈顶的下一层即为上一级的路径。。

知道了这两点后，剩下的就很简单了。用stringstream类，将字符串根据'/'进行分割即可。。

具体代码如下：

class Solution {public:    string simplifyPath(string path) {        stack<string> res_stack;        string res = "";        string token;        istringstream is(path);        while (getline(is, token, '/')) {        if (token == "." or token == "") {        continue;        }        else if (token == ".." and !res_stack.empty()) {        res_stack.pop();        }        else if (token != "..") {        res_stack.push(token);        }        }        while (!res_stack.empty()) {        res = "/" + res_stack.top() + res;        res_stack.pop();        }        if (res == "") {        res = "/";        }        return res;    }};