Given an absolute path for a file (Unix-style), simplify it.
Example
"/home/", => "/home"
"/a/./b/../../c/", => "/c"
Challenge
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
- 將整個路徑按照/分開來
- 用一個棧來存路徑眠蚂，遇到
..時彈出一個，遇到.和空字符串則不變，遇到正常路徑則壓入棧中败徊。
Note
- 如果結(jié)果為空樟结，返回
/ - 彈出棧時要先檢查棧是否為空
public class Solution {
/*
* @param path: the original path
* @return: the simplified path
*/
public String simplifyPath(String path) {
// write your code here
//用stack來存通過split（“ ”）分離的path
//有四種情況驶冒，
// “.”:什么都不做
// “..”：退回上一層
// “”：什么都不做
//其他字符：入棧
Stack<String> stack = new Stack<String>();
String[] strList = path.split("/");
StringBuilder result = new StringBuilder();
for(String curStr : strList) {
if (curStr.equals(".")) {
continue;
} else if (curStr.equals("")) {
continue;
} else if (curStr.equals("..")) {
if (!stack.isEmpty()) {
stack.pop();
continue;
}
} else {
stack.push(curStr);
}
}
//如果為空钦奋，則返回空地址
if (stack.isEmpty()) {
return "/";
}
//不為空蓄喇，從后向前加result
while (!stack.isEmpty()) {
result.insert(0,"/"+stack.pop());
}
return result.toString();
}
}
