Simplify Path
Simplify Path
题目
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”.
思路
建立一个字符串栈，/后面的字符们（直到/为止）作为一个字符串，若字符串为..，若栈不为空，则栈顶元素出栈。若字符串不为”.”或者””,则将字符串入栈。
解题
c++版1
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25class Solution {
public:
string simplifyPath(string path) {
stack<string> ss;
for(int i=0;i<path.size();){
while(i<path.size() && '/'==path[i])
++i;
string s="";
while(i<path.size() && path[i]!='/')
s+=path[i++];
if(".."==s && !ss.empty())
ss.pop();
else if(s!="" && s!="." && s!="..")
ss.push(s);
}
if(ss.empty())
return "/";
string s="";
while(!ss.empty()){
s="/"+ss.top()+s;
ss.pop();
}
return s;
}
};
python版1
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24class Solution:
# @param {string} path
# @return {string}
def simplifyPath(self, path):
ss=[]
i=0
while i<len(path):
while i<len(path) and '/'==path[i]:
i=i+1
s=""
while i<len(path) and path[i]!='/':
s+=path[i]
i=i+1
if(".."==s and len(ss)!=0):
ss.pop()
elif (s!="" and s!="." and s!=".."):
ss.append(s)
if(len(ss)==0):
return "/"
s=""
while(len(ss)!=0):
s="/"+ss[-1]+s
ss.pop()
return s
