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题目描述
Given an absolute path for a file (Unix-style), simplify it.
For example,
path = “/home/”, => “/home”
path = “/a/./b/../../c/”, => “/c”
本题要求对给定的Unix路径进行简化,关键点在于理解Unix路径的规则以及对一些边界条件要加以考虑。
Unix路径规则:
字符 | 含义(英文) | 含义(中文) |
---|---|---|
/ | root directory | 根目录 |
/ | Directory Separator | 路径分隔符 |
. | Current Directory | 当前目录 |
.. | Parent Directory | 上级目录 |
~ | Home Directory | 家目录 |
Corner Cases:
path = “/../” => 输出 “/”.
连续的多个分隔符 ‘/’ , 例如 “/home//foo/” => 忽略重复的分隔符,输出 “/home/foo”.
解题思路
总体来说本题并不难,只要根据unix path规则,细心考虑可能的边界条件即可。
class Solution {
public:
string simplifyPath(string path) {
char slash = '/';
vector<string> dirs;
int begin = 0, end = 0;
int len = path.length();
while (end < len)
{
begin = end;
while (begin < len && path[begin] == slash)
begin++;
end = begin;
while (end < len && path[end] != slash)
end++;
if (begin != end)
{
string tmp(path, begin, end - begin);
if (tmp == ".." && !dirs.empty())
dirs.pop_back();
else if (tmp == ".." && dirs.empty())
continue;
else if (tmp != ".")
dirs.push_back(tmp);
}
}
string simplePath = "";
for (size_t i = 0; i != dirs.size(); ++i)
{
simplePath += slash + dirs[i];
}
if(dirs.empty())
simplePath = slash;
return simplePath;
}
};