Simplify Path -- leetcode

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".

基本思路：

1. 以字符 /  作为分隔符，取子串。

2. 对 .. 作特殊处理。

3. 即不是. 又不是 ..  则存入栈中

4. 最后串接

所遇到的特殊case,

输入          /...      (连着三个小数点）

期望输出  /..      （亦是连着三个小数点）.

进入循环前，在path末尾添加 / ， 作为哨兵，可简化代码。

class Solution {
public:
    string simplifyPath(string path) {
        vector<string> stack;
        string name;
        path.push_back('/');
        for (int i=0; i<path.size(); i++) {
            if (path[i] == '/') {
                if (name == "..") {
                     if (!stack.empty()) stack.pop_back();
                }
                else if (name != "." && !name.empty())
                    stack.push_back(name);

                name.clear();
            }
            else
                name.push_back(path[i]);
        }

        if (stack.empty()) return "/";

        string ans;
        for (auto p: stack)
            ans += "/" + p;

        return ans;
    }
};