LeetCode | Simplify Path

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"

.

题目解析：

根据情况，用栈来做。

例如：

输入1：

/../a/b/c/./..

输出1：

/a/b

模拟整个过程：

1. "/" 根目录

2. ".." 跳转上级目录，上级目录为空，所以依旧处于 "/"

3. "a" 进入子目录a，目前处于 "/a"

4. "b" 进入子目录b，目前处于 "/a/b"

5. "c" 进入子目录c，目前处于 "/a/b/c"

6. "." 当前目录，不操作，仍处于 "/a/b/c"

7. ".." 返回上级目录，最终为 "/a/b"

class Solution {
public:
string simplifyPath(string path) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
stack<string> s;
string str;
for(int i = 0; i < path.size(); i++)
{
if (path[i] == '/')
{
if (str == "..")
{
if (!s.empty())
s.pop();
}
else if (str != "." && str != "")
{
s.push(str);
}

str = "";   //重新清空
}
else
{
str += path[i]; //保存文件名，可能由多个字符组成
}
}

if (str == "..")    //处理最后的情况，
{
if (!s.empty())
s.pop();
}
else if (str != "." && str != "")
s.push(str);

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

string ret;
while(!s.empty())
{
ret = "/" + s.top() + ret;
s.pop();
}

return ret;
}
};