[Leetcode] Simplify Path 化简路径

Simplify Path

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

For example,

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

栈法

复杂度

时间 O(N) 空间 O(N)

思路

思路很简单,先将整个路径按照/分开来,然后用一个栈,遇到..时弹出一个,遇到.和空字符串则不变,遇到正常路径则压入栈中。

注意

  • 如果结果为空,要返回一个/
  • 弹出栈时要先检查栈是否为空

代码

public class Solution {
    public String simplifyPath(String path) {
        Stack<String> stk = new Stack<String>();
        String[] parts = path.split("/");
        for(String part : parts){
            switch(part){
                case ".":
                case "" :
                    break;
                case "..":
                    if(!stk.isEmpty()){
                        stk.pop();
                    }
                    break;
                default:
                    stk.push(part);
            }
        }
        StringBuilder sb = new StringBuilder();
        if(stk.isEmpty()){
            return "/";
        }
        while(!stk.isEmpty()){
            sb.insert(0, "/"+stk.pop());
        }
        return sb.toString();
    }
}

2018/2

class Solution:
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """
        parts = path.split('/')
        simplified = []
        for part in parts:
            if part == '.' or part == '':
                continue
            elif part == '..':
                simplified and simplified.pop()
            else:
                simplified.append(part)
        return '/' + '/'.join(simplified)
    原文作者:ethannnli
    原文地址: https://segmentfault.com/a/1190000003815508
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