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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”.
解析:采用栈的方法。根据文件路径的性质,当出现.的时代表当前目录,可以忽略不计,出现..代表返回上级目录,即代表删除前一个路径。
通过上述的分析,我们可以制定如下的规则: 可以利用栈来存放输出路径
(1) 先去除所有的/,
(2) 当遇见..时弹出栈顶元素,代表删除上一级目录
(3) 遇见字母的时候进行压栈
(4) 将栈中所有元素出栈并加入分割符,此时栈顶元素代表是路径尾部。代码:
public static String simplifypath(String path) {
Stack<String> s = new Stack<>();
String[] p = path.split("/");
for(String str : p) {
if(!s.isEmpty() && str.equals("..")) {
s.pop();
}else if(!str.equals(".") && !str.equals("")&&!str.equals("..")) {
s.push(str);
}
}
List<String> list = new ArrayList<>(s);
return "/"+String.join("/", list);
}