71. Simplify Path Problem Solving Record

Topic description:

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".

Problem solving ideas:

There are three special cases for this question: multiple '/' characters, "../" strings, and "./" strings. Therefore, when traversing, I skip '/' to add other arrays, and then compare the obtained sub-arrays with ".." and ".", and clear the obtained string each time.

Code:

1  class Solution {
 2  public :
 3      string simplifyPath( string path) {
 4          string ret, temp;
 5          int n = path.length();
 6          for ( int i = 0 ; i < n; i++ ) {
 7              if (path [i] == ' / ' ) {
 8                          // Add case 
9 triggered by '/'                  if (temp.empty())
 10                                  // temp is an empty string 
11                      continue ;
 12                  if(temp == " . " ) {
 13                                  // temp is '.' 
14 temp.                      clear();
 15                      continue ;
 16                  }
 17                  if (temp == " .. " ) {
 18                                  // temp is ".." 
19                      int next = ret.rfind( " / " );
 20                      if (next > 0 )
 21                                          // Prevent ret from being an empty string and next is -1 
22                          ret.erase(ret.begin() + next, ret. end());
 23                      else
24                         ret.clear();
25                     temp.clear();
26                 }
27                 else {
28                                 //添加
29                     ret = ret + '/' + temp;
30                     temp.clear();
31                 }
32             }
33             else
34                 temp += path[i];
35         }
36         if (temp == ".." && ret.length()>1) {
37                  // Because '/' is used as the addition condition, the last character may not be '/', so we need to compare again 
38              ret.erase(ret.begin() + ret.rfind( " / " ), ret.end());
 39          }
 40          if (temp != " . " && temp != " .. " && ! temp.empty())
 41              ret = ret + ' / ' + temp;
 42          if ( ret.empty())
 43              ret = " / " ;
 44          return ret;
45     }
46 };