71. Simplify Path 解题记录

Posted 2020-10-31 宵夜

题目描述：

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".

解题思路：

这题的特殊情况有三个：多个‘/‘字符，"../"字符串和"./"字符串。因此我在遍历的时候直接略过‘/‘添加其他数组，再将得到的子数组与".."和"."比较，每次比较清空得到的字符串。

代码：

 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                         //以‘/‘为触发添加情况
 9                 if (temp.empty())
10                                 //temp为空串
11                     continue;
12                 if (temp == ".") {
13                                 //temp为‘.‘
14                     temp.clear();
15                     continue;
16                 }
17                 if (temp == "..") {
18                                 //temp为".."
19                     int next = ret.rfind("/");
20                     if (next > 0)
21                                         //防止ret为空串，next为-1的情况
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                 //因为以‘/‘为添加条件，可能会碰到最后一个字符不是‘/‘的情况，所以最后要再比较一次
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 };