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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; }};
Leetcode: Simplify Path
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