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

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

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".

规律:

1、".."表示跳到上一层目录,删掉它上面挨着的一个路径。

2、"."表示当前目录,直接去掉。

3、如果有多个"/"只保留一个。

4、如果最后有一个"/",去掉不要。

Java:

public class Solution {
public String simplifyPath(String path) {
java.util.LinkedList<String> stack = new java.util.LinkedList<String>();
java.util.Set<String> set = new java.util.HashSet<String>(java.util.Arrays.asList("..", ".", ""));
for (String str : path.split("/")){
if (str.equals("..") && !stack.isEmpty()) stack.pop();
if (!set.contains(str)) stack.push(str);
} String result = "";
for (String str : stack) result = "/" + str + result;
return result.isEmpty() ? "/" : result;
}
}  

Python:

class Solution:
# @param path, a string
# @return a string
def simplifyPath(self, path):
stack, tokens = [], path.split("/")
for token in tokens:
if token == ".." and stack:
stack.pop()
elif token != ".." and token != "." and token:
stack.append(token)
return "/" + "/".join(stack)

C++:

class Solution {
public:
string simplifyPath(string path) {
vector<string> v;
int i = 0;
while (i < path.size()) {
while (path[i] == '/' && i < path.size()) ++i;
if (i == path.size()) break;
int start = i;
while (path[i] != '/' && i < path.size()) ++i;
int end = i - 1;
string s = path.substr(start, end - start + 1);
if (s == "..") {
if (!v.empty()) v.pop_back();
} else if (s != ".") {
v.push_back(s);
}
}
if (v.empty()) return "/";
string res;
for (int i = 0; i < v.size(); ++i) {
res += '/' + v[i];
}
return res;
}
};

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