LeetCode:ZigZag Conversion

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".

刚开始的时候不知道ZigZag Conversion到底是怎么一回事。

看了http://blog.youkuaiyun.com/zhouworld16/article/details/14121477的才知道zigzag有点类似反z的形式排列。

起始一开始考虑的时候，我考虑的nRows都是>=3的。其实并没有考虑nRows=1的情况。导致我写的getRowString()这个函数处于一个无限循环的状态。同时也没有考虑到s为空的时候状况。所以在提交时才导致了Timelimited error。所以才刚开始的地方加上检测语句，当s为空，nRows为1，直接返回字符串s。剩余的情况按照最开始的算法就可以了。

其实在编写过程中，我把row分成3种情况，一种是第一行，一个是最后一行，剩余的是中间行。其实第一行和最后一行一样。比较规律，只要提取间隔为numberOfCircle的位置上的字母。每次取一行，append成一个字符串就好。

---

public class Solution {
     public String convert(String s, int nRows) {
        int length = s.length();
        int index = 1;
        String str = new String();
        if(length == 0 || nRows == 1)return s;
        while(index <= nRows)
        {
            str = str + getRowString(s,index,nRows);
            index = index + 1;
        }
        return str;
    }
   
    public String getRowString(String s,int row,int nRows)
    {
        StringBuilder sb = new StringBuilder();
        int numberOfCircle = 2*nRows - 2;
        int index = 0;
        if(nRows == 1)
        {
            return s;
        }
        if(row == 1)
        {
             while(index<s.length())
            {
                sb.append(s.charAt(index));
                index = index + numberOfCircle;
            }
        }
        else if(row == nRows)
        {
            index = row - 1;
            while(index<s.length())
            {
                sb.append(s.charAt(index));
                index = index + numberOfCircle;
            }
        }
        else
        {
            index = row - 1;
            int interval = 2*(nRows - row);
            while(index<s.length())
            {
                sb.append(s.charAt(index));
                if(index + interval < s.length())
                {
                    sb.append(s.charAt(index + interval));
                }
                index = index + numberOfCircle;
            }
        }
       
        return sb.toString();
    }
}

转载于:https://www.cnblogs.com/jessiading/p/3736683.html