LeetCode 71: Simplify Path

最新推荐文章于 2020-05-11 20:47:16 发布

转载最新推荐文章于 2020-05-11 20:47:16 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/shuashuashua/p/7418395.html

本文介绍了一个简化文件系统路径字符串的Java算法实现。该算法通过栈结构处理输入的路径字符串，去除多余的符号并返回规范化的路径。特别关注了边缘情况处理，如根目录路径和..符号的正确处理。

class Solution {
    public String simplifyPath(String path) {
        if (path == null || path.length() == 0) {
            return path;
        }
        String[] clearStr = path.split("/");
        StringBuilder result = new StringBuilder();
        Stack<String> paths = new Stack<>();
        result.append('/');
        for (String str : clearStr) {
            if (str.equals("") || str.equals(".")) {
                continue;
            }
            
            if (str.equals("..")) {
                if (!paths.isEmpty()) {
                    paths.pop();;
                }
                continue;
            }
            paths.push(str);
        }
        
        while (!paths.isEmpty()) {
            result.insert(1, "/");
            result.insert(1, paths.pop());
        }
        
        return result.length() == 1 ? result.toString() : result.toString().substring(0, result.length() - 1);
    }
}

LOTS of EDGE CASES:

1. Root path, when we return, it needs to be checked.

2. When we meet "..", we shall check whether it is root path or not, but do not forget to skip pushing into stack.

转载于:https://www.cnblogs.com/shuashuashua/p/7418395.html