[leetcode]Simplify Path

问题描写叙述：

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

For example,
path =

"/home/"

, =>

"/home"

path =

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

, =>

"/c"

click to show corner cases.

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"

.

基本思想：

此题能够用一个堆栈存储每层目录或文件。假设两个/之间的内容是".."就弹出栈顶元素。假设遇到“.”或“”就不变。基本思想就是这样。

但要考虑一些corner case。如“/../”,"///"等

代码：

public String simplifyPath(String path) {  //Java
// return "/";
path = path.trim();
if(path.equals("/"))
return "/";

List<String> stack = new LinkedList<String>();

int top = -1;
String tmp = path;
while(!tmp.equals("")){
if(tmp.startsWith("/"))
tmp = tmp.substring(1);
if(tmp.equals(""))
break;

if(tmp.contains("/")){
int pos = tmp.indexOf("/");
String file = tmp.substring(0,pos);
tmp = tmp.substring(pos);
if(file.equals(".")|| file.equals(""))
continue;
if(file.equals("..")){
if(top>-1)
top--;
continue;
}
stack.add(++top,file);

}
else {
String file = tmp;
tmp = "";
if(file.equals("."))
continue;
if(file.equals("..")){
if(top>-1)
top--;
continue;
}
stack.add(++top,file);

}
}
//generate path;
String rpath = "/";
for(int i = 0; i <=top ; i++)
rpath += stack.get(i)+"/";
if(top >=0)
rpath = rpath.substring(0,rpath.length()-1);
return rpath;

}