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Saturday, January 5, 2013

[LeetCode] Simplify Path 解题报告


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".
» Solve this problem

[解题思路]
利用栈的特性,如果sub string element
1. 等于“/”,跳过,直接开始寻找下一个element
2. 等于“.”,什么都不需要干,直接开始寻找下一个element
3. 等于“..”,弹出栈顶元素,寻找下一个element
4. 等于其他,插入当前elemnt为新的栈顶,寻找下一个element

最后,再根据栈的内容,重新拼path。这样可以避免处理连续多个“/”的问题。


[Code]
1:       string simplifyPath(string path) {   
2:            // Start typing your C/C++ solution below   
3:            // DO NOT write int main() function   
4:            vector<string> stack;   
5:            assert(path[0]=='/');   
6:            int i=0;   
7:            while(i< path.size())   
8:            {   
9:                 while(path[i] =='/' && i< path.size()) i++; //skip the begining '////'  
10:                 if(i == path.size())   
11:                      break;   
12:                 int start = i;   
13:                 while(path[i]!='/' && i< path.size()) i++; //decide the end boundary  
14:                 int end = i-1;   
15:                 string element = path.substr(start, end-start+1);   
16:                 if(element == "..")   
17:                 {   
18:                      if(stack.size() >0)   
19:                      stack.pop_back();   
20:                 }   
21:                 else if(element!=".")   
22:                 stack.push_back(element);      
23:            }   
24:            if(stack.size() ==0) return "/";   
25:            string simpPath;   
26:            for(int i =0; i<stack.size(); i++)   
27:            simpPath += "/" + stack[i];   
28:            return simpPath;   
29:       }   




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