string leetcode-6.ZigZag

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 {
 public:
     string convert(string s, int numRows) {
         int len = s.length();
         if(len < )
             return s;
         char ans[len][len];
　　　　　　 //记忆化数组初始化为'#'
         for (int i = ; i < len; i++)
         {
             for (int j = ; j < len; j++)
             {
                 ans[i][j] = '#';
             }
         }
         int i = , col = ;
         while (i < len)
         {
　　　　　　　　　//竖列
             for (int j = ; j < numRows; j++)
             {
                 ans[j][col] = s[i++];
                 if (i >= len)
                     break;
             }
　　　　　　　　  //斜列
             for (int j = numRows - ; j >= ; j--)
             {
                 if (i >= len)
                     break;
                 ans[j][++col] = s[i++];
             }
             col++;
             if (i >= len)
                 break;
         }
         s = "";
　　　　　　 //行遍历，得结果
         for (int i = ; i < len; i++)
         {
             for (int j = ; j < len; j++)
             {
                 if (ans[i][j] != '#')
                     s += ans[i][j];
             }
         }
         return s;
     }
 };

 //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 {
 public:
     string convert(string s, int numRows) {
         if(numRows == )
             return s;
         int len = s.length();
         vector<string> rows(min(numRows, len));
         bool change = false;
         int currow = ;
　　　　　　 //C++ 11的写法
         for(char c : s)
         {
             rows[currow] += c;
             if(currow ==  || currow == numRows-)
                 change = !change;
             currow += (change) ?  : -;
         }
         s = "";
         for(string per : rows)
             s += per;

         return s;
     }
 };
//真的