这道题其实有两个思路

思路一：按顺序把每个字母分配到对应的行，再把行联接起来。具体是用一个string array来代表每一行，用一个current_row来记住现在的行数，然后用go记住方向，向下就是+1，向上就是-1，这样可以不停的运算更新current_row， 在顶行和底行反转go。

class Solution:    def convert(self, s, numRows):        """        :type s: str        :type numRows: int        :rtype: str        """        if len(s) == 0 or numRows <= 1:            return s                row_list = ["" for x in range(numRows)]        current_row = 0        go = -1                for c in s:            row_list[current_row] += c            if (current_row == 0 or current_row == numRows-1):                go = -go            current_row += go                    rst = ""            for row in row_list:            rst += row        return rst

 

思路二：就是直接算出下一个字母是哪一个。size是numRows-2，between就是numRows+size = 2*numRows-2 

class Solution {    public String convert(String s, int numRows) {        if (numRows <= 1) {            return s;        }        StringBuilder rst = new StringBuilder();        int size = 2 * numRows - 2;          for (int i = 0; i < numRows; i++) {            for (int j = i; j < s.length(); j += size) {                rst.append(s.charAt(j));                int between = j + size - 2 * i;                if(i != 0 && i != numRows - 1 && between < s.length()){
   //except the first row and the last row                   rst.append(s.charAt(between));                 }             }                           }           return rst.toString();      }}

 

个人觉得其实第二个方法比较不好像，第一个方法反而比较直观。