LeetCode刷题笔录Simplify Path-优快云博客

本文深入探讨了如何将Unix风格的绝对路径进行简化的过程，通过使用栈数据结构，遵循特定规则处理路径中的符号，最终实现路径的简洁化。包括对常见特例的处理，如连续的斜杠和特殊符号'..'、'.'的含义与作用。

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".

这题需要用到一个stack，利用"/"来分割出每一个substring。"/"本身则不做处理直接跳过，再最后构造结果string的时候再按照规则加上/，这样就可以不用管连续有很多个/的情况。

substring处理如下：

1.如果是"."，什么都不做

2.如果是".."，从stack里pop出一个元素

3.如果是一般字符串，push进stack

最后把stack里的元素按照栈底到栈顶的顺序接在一起就行了。

public class Solution {
    public String simplifyPath(String path) {
        if(path == null || path.length() == 0)
            return "";
        LinkedList<String> stack = new LinkedList<String>();

        char[] arr = path.toCharArray();
        int i = 0;
        while(i < arr.length){
            if(arr[i] == '/'){
                i++;
                continue;
            }
            int start = i;
            int end = start;
            while(end < arr.length && arr[end] != '/')
                end++;
            i = end;
            String sub = path.substring(start, end);
            
            if(sub.equals(".")){
                continue;
            }
            else if(sub.equals("..")){
                if(!stack.isEmpty())
                    stack.remove(stack.size() - 1);
            }
            else{
                stack.add(sub);
            }
        }
        
        StringBuffer str = new StringBuffer("/");
        if(stack.isEmpty())
            return str.toString();
        while(!stack.isEmpty()){
            str.append(stack.pollFirst()).append("/");
        }
        str.deleteCharAt(str.length() - 1);
        return str.toString();
    }
}