71 Simplify Path 简化路径
Description:
Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.
In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level.
Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.
Example:
Example 1:
Input: "/home/"
Output: "/home"
Explanation: Note that there is no trailing slash after the last directory name.
Example 2:
Input: "/../"
Output: "/"
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:
Input: "/home//foo/"
Output: "/home/foo"
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Example 4:
Input: "/a/./b/../../c/"
Output: "/c"
Example 5:
Input: "/a/../../b/../c//.//"
Output: "/c"
Example 6:
Input: "/a//b////c/d//././/.."
Output: "/a/b/c"
题目描述:
以 Unix 风格给出一个文件的绝对路径，你需要简化它。或者换句话说，将其转换为规范路径。
在 Unix 风格的文件系统中，一个点（.）表示当前目录本身；此外，两个点 （..） 表示将目录切换到上一级（指向父目录）；两者都可以是复杂相对路径的组成部分。更多信息请参阅：Linux / Unix中的绝对路径 vs 相对路径
请注意，返回的规范路径必须始终以斜杠 / 开头，并且两个目录名之间必须只有一个斜杠 /。最后一个目录名（如果存在）不能以 / 结尾。此外，规范路径必须是表示绝对路径的最短字符串。
示例 :
示例 1：
输入："/home/"
输出："/home"
解释：注意，最后一个目录名后面没有斜杠。
示例 2：
输入："/../"
输出："/"
解释：从根目录向上一级是不可行的，因为根是你可以到达的最高级。
示例 3：
输入："/home//foo/"
输出："/home/foo"
解释：在规范路径中，多个连续斜杠需要用一个斜杠替换。
示例 4：
输入："/a/./b/../../c/"
输出："/c"
示例 5：
输入："/a/../../b/../c//.//"
输出："/c"
示例 6：
输入："/a//b////c/d//././/.."
输出："/a/b/c"
思路:
利用栈存储
以 '/'作为分隔符插入输入字符, 碰到 '..'且栈非空就弹出
碰到 '.'就跳过
其余的插入
最后将栈中的元素插入结果
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
string simplifyPath(string path)
{
vector temp;
string result;
istringstream iss(path);
while (getline(iss, result, '/'))
{
if (!result.empty() and result != "." and result != "..") temp.push_back(result);
else if (!temp.empty() and result == "..") temp.pop_back();
}
result.clear();
if (temp.empty()) result = "/";
for (string &s: temp) result += "/" + s;
return result;
}
};
Java:
class Solution {
public String simplifyPath(String path) {
String paths[] = path.split("/");
List temp = new ArrayList<>();
for (String s : paths) {
if (!s.isEmpty() && !".".equals(s) && !"..".equals(s)) temp.add(s);
else if (!temp.isEmpty() && "..".equals(s)) temp.remove(temp.size() - 1);
}
if (temp.isEmpty()) return "/";
return "/" + String.join("/", temp);
}
}
Python:
class Solution:
def simplifyPath(self, path: str) -> str:
return "/" + "/".join(functools.reduce(lambda x, y: x[:-1] if y == ".." else x + [y] if y and y != "." else x, path.split("/"), []))