LeetCodeOJ.ZigZag Conversion

试题请参见: https://leetcode.com/problems/zigzag-conversion/

题目概述

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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

解题思路

解释一下什么叫ZigZag, 一开始误解了题意折腾了好久.

可能我们需要用更多的行数可以使这个问题变得更明朗一些:

0       8       H
1     7 A     G I
2   6   B   F   J
3 5     C E     K
4       D       L

对于如上字符串的输入是0123456789ABC...L, 输出是08H17AGI26B...

对于如上例子, 我的思路是把字符串3组, 其中(0 .. 7)为一组, (8.. G)为一组, (H.. L)为一组. 除最后一组外, 每组包含numRows * 2 - 2个元素(其中numRows != 1)。

那么每组中每一个元素应该位于第几行呢? 我们以(0.. 7)这组数据为例:

对于(0 .. 4), 所在行号等于字符串的索引ind % (numRows * 2 - 2)，
对于(5 .. 7), 所在行号等于numRows * 2 - 2 - rowNumber, 其中rowNumber = ind % (numRows * 2 - 2). (你会发现这一组中的每一行索引值相加是定值, 1 + 7 == 2 + 6 == 3 + 5 == numRows * 2 - 2)

源代码

public class Solution {
    public String convert(String s, int numRows) {
        StringBuilder[] rowStrings = new StringBuilder[numRows];
        for ( int i = 0; i < numRows; ++ i ) {
            rowStrings[i] = new StringBuilder();
        }

        for ( int i = 0; i < s.length(); ++ i ) {
            char currentChar = s.charAt(i);
            int rowNumber = numRows == 1 ? 0 : i % (numRows * 2 - 2);

            if ( rowNumber < numRows ) {
                rowStrings[rowNumber].append(currentChar);
            } else {
                rowStrings[numRows * 2 - 2 - rowNumber].append(currentChar);
            }
        }

        StringBuilder zigZag = new StringBuilder();
        for ( int i = 0; i < numRows; ++ i ) {
            zigZag.append(rowStrings[i]);
        }
        return zigZag.toString();
    }

    public static void main(String[] args) {
        Solution s = new Solution();

        System.out.println(s.convert("A", 1));                  // A
        System.out.println(s.convert("ABC", 2));                // ACB
        System.out.println(s.convert("ABCD", 2));               // ACBD
        System.out.println(s.convert("PAYPALISHIRING", 3));     // PAHNAPLSIIGYIR
    }
}