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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”.
题目大意:给定1个绝对路径,简化路径
注意1下几点:
1、“..”表示跳到上1层目录
2、“.”表示当前目录
3、“////”多个/当1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;//寄存目录
path+="/";//避免处理边界,即/a这类情况
string temp = "";//用作临时string
for(int i = 0 ; i< path.length() ;)
{
if(path[i] == '/')
{
if(temp=="..") {//返回上1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;
}
};
上一篇 UML思维导图