题目描述：之字打印出字符串，并返回结果。

题目链接：Leetcode 6. ZigZag Conversion

思路就是初始化n行字符串，然后像搜索一样一个上一个下去搜索。

代码如下

class Solution(object):
    def convert(self, s, numRows):
        """ :type s: str :type numRows: int :rtype: str """
        if numRows == 1 or numRows >= len(s):
            return s

        L = [''] * numRows
        index, step = 0, 1

        for x in s:
            L[index] += x
            if index == 0:
                step = 1
            elif index == numRows -1:
                step = -1
            index += step

        return ''.join(L)

class Solution {
    public String convert(String s, int numRows) {
        if (numRows == 1 || s.length() <= numRows) {
            return s;
        }
        StringBuffer[] ans = new StringBuffer[numRows];  // 初始化n行 然后随着之字走来添加
        
        int index = 0, step = 1;
        char[] chars = s.toCharArray();
        for (char c : chars) {
            if (ans[index] == null) {
                ans[index] = new StringBuffer();
            }
            ans[index].append(c);
            if (index == 0) {
                step = 1;
            } else if (index == (numRows - 1)) {
                step = -1;
            }
            index += step;
        }
        StringBuffer sb = new StringBuffer();
        for (StringBuffer rs : ans) {
            sb.append(rs);
        }
        return sb.toString();
    }
    
}

参考链接