[LeetCode][Java] ZigZag Conversion

题目:

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".

题意:

按照给定的格式,把字符串按列重新排列,之后按行返回新的字符串。

算法分析:

我们将字符在字符串中的序号拿出进行排列分析

*  一: 2排的时候,1到n的排序

*  1 3 5 7 9  。。。

*  2 4 6 8 10 。。。

*  二: 3排的时候,1到n的排序

*  1     5     9  。。。

*  2 4  6 8 10 。。。

*  3     7     11 。。。

*  三: 4排的时候,1到n的排序

*  1       7          13 。。。

*  2   6  8    12 14 。。。

*  3 5   9  11    15 。。。

*  4     10          16 。。。

*  找到规律没有?如果没有找到, 你可以继续写5排的情况。很快你就可以找到规律。这是一个解决问题的方法。

*  当我们遇到难缠的问题的时候,我们先考虑简单的情形,看看能不能找到规律。这个题目,我们通过写出来这些特殊情况。

*  我们发现如下规律,这里我们假设我们分成m排:

 *  1 第i排从i开始

 *  2 第i排两个数的间隔是2(i-1),2(m-i)交替

代码如下:

public class Solution
{
    public String convert(String s, int numRows)
    {
        String result = "";

        if(numRows == 1)
        {
          return s;
        }

        for(int i = 0; i < numRows; i++)
        {
          int j = i;
          boolean flag = true;//通过flag控制两个间隔的切换
          while(j < s.length())
          {
            result+=(s.charAt(j));

            if(i == 0 || i == numRows - 1) //开始和结尾行的两个数之间的间隔都是2(i-1)
              j += 2 * (numRows - 1);
            else
            {
              if(flag)
              {
                j += 2 * (numRows - 1 - i);
                flag = false;
              }
              else
              {
                j += 2 * i;
                flag = true;
              }
            }
          }
        }
        return result;
    }
}

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时间: 2024-10-13 15:08:08

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