[leetcode] ZigZag Conversion *

 1 /*
 2  * Just find the rule; Do not forget to check
 3  * special cases (e.g. nRows = 1 ...)
 4  *
 5  * @author: HZT
 6  * @date: 2013-3-9
 7  */
 8 
 9 #include <iostream>
10 #include <string>
11 using namespace std;
12 
13 class Solution {
14 public:
15     string convert(string s, int nRows) {
16         // Start typing your C/C++ solution below
17         // DO NOT write int main() function
18 
19         // DO NOT forget this!!
20         if(nRows == 1) return s;
21         if(s.empty()) return s;
22         int len = s.length();
23 
24         char* ans = new char[len+1];
25 
26         int idx, ansIdx=0;
27         int delta = 2 * nRows - 2;
28         // each row
29         for(int r=1; r<=nRows; r++){
30             idx = r;
31             while(idx <= len){
32                 ans[ansIdx++] = s[idx-1];
33                 if(r!=1 && r!=nRows && 2*nRows-2*r+idx <= len){
34                     ans[ansIdx++] = s[2*nRows-2*r+idx-1];
35                 }
36 
37                 idx += delta;
38             }
39         }
40 
41         ans[len] = '\0';
42         return ans;
43     }
44 };
45 
46 int main(){
47     string s = "PAYPALISHIRING";
48     int nRows = 3;
49 
50     //string s = "A";
51     //int nRows = 1;
52 
53     cout << (new Solution)->convert(s, 3) << endl;
54 
55     return 0;
56 
57 }