6. ZigZag Conversion

题面

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

样例

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I
 //思路1：
 //完全按照样例来模拟过程
 //时间：O(n2)
 //空间：O(n2)
 1 class Solution {
 2 public:
 3     string convert(string s, int numRows) {
 4         int len = s.length();
 5         if(len < 3)
 6             return s; 
 7         char ans[len][len];
　　　　　　 //记忆化数组初始化为'#'
 8         for (int i = 0; i < len; i++)
 9         {
10             for (int j = 0; j < len; j++)
11             {
12                 ans[i][j] = '#';
13             }
14         }
15         int i = 0, col = 0;
16         while (i < len)
17         {
　　　　　　　　　//竖列
18             for (int j = 0; j < numRows; j++)
19             {
20                 ans[j][col] = s[i++];
21                 if (i >= len)
22                     break;
23             }
　　　　　　　　  //斜列
24             for (int j = numRows - 2; j >= 1; j--)
25             {
26                 if (i >= len)
27                     break;
28                 ans[j][++col] = s[i++];
29             }
30             col++;
31             if (i >= len)
32                 break;
33         }
34         s = "";
　　　　　　 //行遍历，得结果
35         for (int i = 0; i < len; i++)
36         {
37             for (int j = 0; j < len; j++)
38             {
39                 if (ans[i][j] != '#')
40                     s += ans[i][j];
41             }
42         }
43         return s;
44     }
45 };
 //思路2：官方题解
 //1. 把每一行当作一个字符串进行处理，vector<string> 当作容器；
 //2. 遍历字符串，为各行字符串添加元素；
 //3. 行序号的变换（+1/-1）依靠一个bool变量控制，每当到达第一行/最后一行，bool变量取反 
 //时间：O(n)
 //空间：O(n)
 1 class Solution {
 2 public:
 3     string convert(string s, int numRows) {
 4         if(numRows == 1)
 5             return s;
 6         int len = s.length();
 7         vector<string> rows(min(numRows, len));
 8         bool change = false;
 9         int currow = 0;
　　　　　　 //C++ 11的写法
10         for(char c : s)
11         {
12             rows[currow] += c;
13             if(currow == 0 || currow == numRows-1)
14                 change = !change;
15             currow += (change) ? 1 : -1;
16         }
17         s = "";
18         for(string per : rows)
19             s += per;
20         
21         return s;
22     }
23 };
//真的🐮！