题目链接:Simplify Path
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”.
这道题的要求是简化一个Unix风格下的文件的绝对路径。
字符串处理,由于”..”是返回上级目录(如果是根目录则不处理),因此可以考虑用栈记录路径名,以便于处理。需要注意几个细节:
- 重复连续出现的’/’,只按1个处理,即跳过重复连续出现的’/’;
- 如果路径名是”.”,则不处理;
- 如果路径名是”..”,则需要弹栈,如果栈为空,则不做处理;
- 如果路径名为其他字符串,入栈。
最后,再逐个取出栈中元素(即已保存的路径名),用’/’分隔并连接起来,不过要注意顺序呦。
时间复杂度:O(n)
空间复杂度:O(n)
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class Solution
{
public:
string simplifyPath(string path)
{
stack<string> ss; // 记录路径名
for(int i = 0; i < path.size(); )
{
// 跳过斜线'/'
while(i < path.size() && '/' == path[i])
++ i;
// 记录路径名
string s = "";
while(i < path.size() && path[i] != '/')
s += path[i ++];
// 如果是".."则需要弹栈,否则入栈
if(".." == s && !ss.empty())
ss.pop();
else if(s != "" && s != "." && s != "..")
ss.push(s);
}
// 如果栈为空,说明为根目录,只有斜线'/'
if(ss.empty())
return "/";
// 逐个连接栈里的路径名
string s = "";
while(!ss.empty())
{
s = "/" + ss.top() + s;
ss.pop();
}
return s;
}
};