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LeetCode 071 Simplify Path

2015-10-06 15:47 351 查看

题目描述

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”.

分析

“.”表示当前路径,不打印

“..”表示上级路径,其左边第一个有效路径略过

可以先用“/”分隔Path,然后用栈操作,最后组合结果

代码

[code]    public static String simplifyPath(String path) {

        if (path == null) {
            return null;
        }

        String[] names = path.split("/");

        int eat = 0;

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

        for (int i = names.length - 1; i >= 0; i--) {

            String token = names[i];

            // token是"..", 表示上级路径,前一个路径不打印
            // token是".", 表示当前路径,自身不打印
            // token是"", 表示为两个"/"相连,不做操作
            // eat>0,表示左边不打印
            // 否则,将token入栈
            if (token.equals("..")) {
                eat++;
            } else if (token.equals(".")) {
                // do nothing
            } else if (token.equals("")) {
                // do nothing
            } else {

                if (eat > 0) {
                    eat--;
                } else {
                    stack.push(token);
                }
            }
        }

        StringBuilder s = new StringBuilder();

        s.append("/");

        // 最后一个不加"/",所以while判断条件是>1
        while (stack.size() > 1) {
            s.append(stack.pop());
            s.append("/");
        }

        // 最后一个不加"/"
        if (!stack.isEmpty()) {
            s.append(stack.pop());
        }

        return s.toString();
    }


参考

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