[LintCode] Simplify Path 简化路径

Given an absolute path for a file (Unix-style), simplify it.

Have you met this question in a real interview?

Example

"/home/"

, =>

"/home"

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

, =>

"/c"

Challenge

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"

.

LeetCode上的原题，请参见我之前的博客Simplify Path。

解法一：

class Solution {
public:
/**
* @param path the original path
* @return the simplified path
*/
string simplifyPath(string& path) {
int left = 0, right = 0, n = path.size();
stack<string> s;
string res = "";
while (right < n) {
while (left < n && path[left] == '/') ++left;
right = left;
while (right < n && path[right] != '/') ++right;
string t = path.substr(left, right - left);
if (t == "..") {
if (!s.empty()) s.pop();
} else if (t != ".") {
if (!t.empty()) s.push(t);
}
left = right;
}
while (!s.empty()) {
res = "/" + s.top() + res; s.pop();
}
return res.empty() ? "/" : res;
}
};

解法二：

class Solution {
public:
/**
* @param path the original path
* @return the simplified path
*/
string simplifyPath(string& path) {
string res, t;
stringstream ss(path);
vector<string> v;
while (getline(ss, t, '/')) {
if (t == "" || t == ".") continue;
if (t == ".." && !v.empty()) v.pop_back();
else if (t != "..") v.push_back(t);
}
for (string s : v) res += "/" + s;
return res.empty() ? "/" : res;
}
};