LeetCode:Simplify Path

题目链接

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"

.

分析：需要注意的是/…可以表示名字为…的路径，路径的最后可能没有/。可以利用栈，碰到正常路径压入栈中，碰到/.不作任何操作，碰到/..删除栈顶元素。下面代码中用数组来模拟栈 本文地址

class Solution {
public:
string simplifyPath(string path) {
int len = path.size();
vector<string> vec;
int i = 0, index = 0;
while(i < len)
{
int j = path.find('//', i + 1);
string tmp;
if(j != string::npos)
tmp = path.substr(i, j - i);
else {tmp = path.substr(i, len); j = len;}

if(tmp == "/");
else if(tmp == "/.");
else if(tmp == "/..")
{if(!vec.empty())vec.pop_back();}
else
vec.push_back(tmp);
i = j;
}
if(vec.empty())return "/";
else
{
string res;
for(int i = 0; i < vec.size(); i++)
res += vec[i];
return res;
}
}
};

【版权声明】转载请注明出处：/article/4879692.html