[LeetCode] 6. ZigZag Conversion

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。