Leetcode: Simplify Path

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"

.

分析：细节题。在考虑各cornoer cases的基础上完成所需要求即可。在解此题前要有一个大致的执行逻辑，比如如何执行"."和".."，以及将"filename"还是将"filename/"增加到当前结果中。在下面的代码中，如果是"."我们不做操作；如果是"filename"则将"filename/"加到result末尾；如果是“..”，我们先pop出result末尾的'/'（因为我们遇到"filename"是将"filename/"加到result中），然后再将filename从result中pop。这样做逻辑比较清晰，不易出错，刚开始，我的做法是先将"/"加入到result中，然后在分情况处理".",".."以及"filename"的情况，但一直有wrong answer。

class Solution {
public:
string simplifyPath(string path) {
string result;
int n = path.length();
//two corner cases
if(n == 0) return result;
if(path == "/../") return "/";
//push root directory
result.push_back('/');
for(int i = 0; i < n;){
while(i < n && path[i] == '/')i++;//skip redundant '/'
int j = i;
while(j < n && path[j] != '/')j++;//find next '/'

if(i == n) break;

string file = path.substr(i,j-i);//get file name
//three cases
if(file == ".."){
if(result != "/"){
result.pop_back();//pop trailing '/'
size_t pos = result.find_last_of('/');
result = result.substr(0,pos+1);
}
}else if(file != "."){
result += file + "/";
}
i = j;
}
if(result.length() > 1 && result.back() == '/') result.pop_back();//pop trailing '/'
return result;
}
};