[Leetcode] Decode String, Solution

Given an encoded string, return it's decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

[Thoughts]
凡是这种decode的题，基本上都是用栈。当遇到左括号的时候，保留住当前的栈状态，当遇到右括号的时候，弹出当前栈状态，做一番处理。具体看code



[Code]

1:    string decodeString(string s) {  
2:      stack<int> repeat_counts;  
3:      stack<string> pre_fix;  
4:        
5:      int num = 0;  
6:      string decoded = "";  
7:      string cur = "";  
8:        
9:      s = "1[" + s +"]"; // wrap the input into a bracket to simplify the logic  
10:      for(int i = 0; i< s.size(); i++) {  
11:        if(isdigit(s[i])) {  
12:          num = num*10 + s[i] -'0';  
13:          continue;  
14:        }  
15:          
16:        if(s[i] == '[') {  
17:          repeat_counts.push(num);  
18:          num = 0;  
19:          pre_fix.push(cur);  
20:          cur = "";  
21:          continue;  
22:        }  
23:          
24:        if(s[i] == ']') {  
25:          int repeat = repeat_counts.top();  
26:          repeat_counts.pop();  
27:          string prefix = pre_fix.top();  
28:          pre_fix.pop();  
29:            
30:          cur = prefix + repeatString(cur, repeat);  
31:          continue;  
32:        }  
33:          
34:        cur += s[i];  
35:      }  
36:        
37:      return cur;  
38:    }  
39:      
40:    string repeatString(string& organic, int repeats) {  
41:      string result = "";  
42:      for(int i =0; i< repeats; i++) {  
43:        result += organic;  
44:      }  
45:        
46:      return result;  
47:    }