[leetcode]Simplify Path

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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".

演算法思路：

棧。設定兩指標，根據‘/’分割出一個一個的路徑名，壓棧。pop的時候，遇到‘.’和“”（//生產的）不用管，".."復原，並記錄復原次數。

注意：當路徑為空白時，要返回 / 。

吐槽：有一個奇葩case -> /... 後來才反映過來，人家的路徑名叫...搞笑嗎？

 1 public class Solution { 2     public String simplifyPath(String path) { 3         if(path == null || path.length() == 0) return ""; 4         int start = 0; 5         Stack<String> stack = new Stack<String>(); 6         for(int i = 1; i < path.length(); i++){ 7             if(path.charAt(i) == ‘/‘ || i == path.length() - 1){ 8                    String s = (path.charAt(i) == ‘/‘) ? path.substring(start + 1, i) : path.substring(start + 1, i + 1); 9                    stack.push(s);10                    start = i;11             }12         }13         StringBuilder sb = new StringBuilder();14         int traceBack = 0;15         while(!stack.isEmpty()){16             String str = stack.pop();17             if(".".equals(str) || str.length() == 0)continue;18             else if("..".equals(str)) {19                 traceBack++;20             }else{21                 traceBack--;22                 if(traceBack < 0){23                     sb.insert(0,  "/" + str);24                     traceBack = 0;25                 }26             }27         }28         return sb.toString().length() == 0 ? "/" : sb.toString();29     }30 }