【一天一道LeetCode】#71. Simplify Path
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(一)题目
Given an absolute path for a file (Unix-style), simplify it.
For example,
path = “/home/”, => “/home”
path = “/a/./b/../../c/”, => “/c”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”.
(二)解题
题目大意:给定一个绝对路径,简化路径
注意一下几点:
1、“..”表示跳到上一层目录
2、“.”表示当前目录
3、“////”多个/当一个处理
在这里,我们可以利用stack来处理,“..”则把栈顶元素弹出,“.”则不作处理,其他的表示目录的名字就直接压栈。
具体思路可以看代码:
class Solution {public: string simplifyPath(string path) { stack<string> pathStack;//存放目录 path+="/";//避免处理边界,即/a这种情况 string temp = "";//用作临时string for(int i = 0 ; i< path.length() ;) { if(path[i] == '/') { if(temp=="..") {//返回上一层目录 if(!pathStack.empty()) pathStack.pop();//如果非空则弹出,为空就不做处理 } else if(temp=="."|| temp =="") {}//不做处理,当前目录 else pathStack.push(temp);//正常的目录 while(i<path.length()&&path[i]=='/') i++;//跳过连续的‘/’ temp = "";//temp初始化 } else { temp+=path[i++]; } } string ret; if(pathStack.empty()) ret = "/";//为空时直接输出 else { while(!pathStack.empty())//非空则按格式输出 { string eachPath = pathStack.top(); pathStack.pop(); ret = "/"+eachPath + ret; } } return ret; }};
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