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[LeetCode#71]Simplify Path

时间:2015-09-03 07:00:34      阅读:158      评论:0      收藏:0      [点我收藏+]

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Problem:

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

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

Analysis:

This problem is very very easy, if you really understand how to read linux path. 
Big idea:
Maintain a stack, then interpret the element between "/" one by one.
iff the element is "..", we pop one element out
iff the element is ".", we do nothing with it.
iff the element is "directory/file name", we push it onto the stack. 

Initial Wrong Solution:

public class Solution {
    public String simplifyPath(String path) {
        if (path == null)
            throw new IllegalArgumentException("The passed in path is null!");
        String[] stubs = path.split("/");
        Stack<String> stack = new Stack<String> ();
        for (String stub : stubs) {
            if (stub.equals("..")) {
                if (!stack.isEmpty())
                    stack.pop();
            } else if (!stub.equals(".")) {
                stack.push(stub);
            } 
        }
        StringBuffer ret = new StringBuffer();
        while (!stack.isEmpty()) {
            ret.insert(0, stack.pop());
            ret.insert(0, "/");
        }
        if (ret.charAt(0) != ‘/‘)
            ret.insert(0, "/");
        return ret.toString();
    }
}

Mistakes Analysis:

Even though it is easy, I have made following mistakes in implementation.
Mistake 1: ignore the case "/", which would result in "no element pushed onto the stack".
Then tried following error fix:
if (ret.charAt(0) != ‘/‘)
    ret.insert(0, "/");
This is absolutely wrong, since ret is empety, this would cause expcetion.
Fix: 
if (ret.length() == 0)
    return "/";

Mistake 2: ignore the String.spilit could produce "" as element.
Case: "/..."
String.split => ["", "..."]
Fix:
for (String stub : stubs) {
    if (stub.equals("..")) {
    ...
    }
}

Solution:

public class Solution {
    public String simplifyPath(String path) {
        if (path == null)
            throw new IllegalArgumentException("The passed in path is null!");
        String[] stubs = path.split("/");
        Stack<String> stack = new Stack<String> ();
        for (String stub : stubs) {
            if (stub.equals(""))
                continue;
            if (stub.equals("..")) {
                if (!stack.isEmpty())
                    stack.pop();
            } else if (!stub.equals(".")) {
                stack.push(stub);
            } 
        }
        StringBuffer ret = new StringBuffer();
        while (!stack.isEmpty()) {
            ret.insert(0, stack.pop());
            ret.insert(0, "/");
        }
        if (ret.length() == 0)
            return "/";
        return ret.toString();
    }
}

[LeetCode#71]Simplify Path

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原文地址:http://www.cnblogs.com/airwindow/p/4779850.html

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