LeetCode :6. ZigZag Conversion

LeetCode : ZigZag Conversion

题目:
LeetCode:6. ZigZag Conversion

描述：

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)

举例：

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".  

分析：

题目的意思是把字符串上下上下走之字形状,然后按行输出.

0   4    8
1 3 5  7 9
2   6    10

输出0->4->8->1->3->5...->6->10

1. 根据其数组的读写规律，从首行到末行进行依次写入操作
2. 从首行到末行，根据其递增的特性，从左到右写入

• 规律如下：
  1.1 首行与末行 每一列递增规律为nInterval*(nColumn ) + (i - 1)
  1.2 中间行规律为nInterval * ((nColumn + 1) / 2) + (i - 1) * (nColumn % 2 ? -1 : 1) 0 + 1 = 1
  2.1 首行每个元素间隔为 nInterval，中间行 间隔为 nInterval - 2 * i 与 2 * i交替,尾行 间隔为 nInterval.
  备注：其中nInterval是按照规律读写同一位置的间隔。nInterval = (numRows << 1) - 2， i 为行数，ncolumn当前列数，numRows为指定每列数量。

代码：

{
string convertEx(string s, int numRows) {
    int nLen = s.size();
    if (1 == nLen || 1 == numRows)
    {
        return s;
    }
    string strRes(nLen, ' ');
    int nRow = 0;
    int nColumn = 0;
    int nRule = 0;
    int nInterval = (numRows << 1) - 2; // 移位代替 * 2
    int nIndex = 0;
    /*
    首行：     nRule = interval*(nColumn) + nRow
    中间行：   nRule = interval * ((nColumn + 1) / 2) + nRow * (nColumn % 2 ? -1 : 1)  0 + 1 = 1
    尾行：     nRule = interval*(nColumn) + nRow
    */
    while (nRow < numRows)
    {
        if (0 == nRow)
        {
            nColumn = 0;
            nRule = nInterval * nColumn;
            while (nRule < nLen)
            {
                strRes[nIndex] = s[nRule];
                ++nColumn;
                ++nIndex;
                nRule = nInterval * nColumn;
            }
        }
        else if (nRow < numRows - 1)
        {
            nColumn = 0;
            nRule = nInterval * ((nColumn + 1) >> 1) + nRow * (nColumn % 2 ? -1 : 1);
            while (nRule < nLen)
            {
                strRes[nIndex] = s[nRule];
                ++nColumn;
                ++nIndex;
                nRule = nInterval * ((nColumn + 1) >> 1) + nRow * (nColumn % 2 ? -1 : 1);
            }
        }
        else
        {
            nColumn = 0;
            nRule = nInterval * nColumn + nRow;
            while (nRule < nLen)
            {
                strRes[nIndex] = s[nRule];
                ++nColumn;
                ++nIndex;
                nRule = nInterval * nColumn + nRow;
            }
        }
        ++nRow;
    }
    return strRes;
}

}

string convert(string s, int numRows) {
    int nLen = s.size();
    if (1 == nLen || 1 == numRows)
    {
        return s;
    }
    string strRes(nLen, ' ');
    int nInterval = (numRows << 1) - 2; // 移位代替 * 2
    int nIndex = 0;
    // 首行 间隔为 nInterval
    for (int i = 0; i < nLen; i += nInterval)
    {
        strRes[nIndex++] = s[i];
    }
    // 中间行 间隔为  nInterval - 2 * i 与 2 * i交替
    for (int i = 1; i < numRows - 1; i++)
    {
        int nInterTemp = i << 1;
        for (int j = i; j < nLen; j += nInterTemp)
        {
            strRes[nIndex++] = s[j];
            nInterTemp = nInterval - nInterTemp; // 和为nInterval 互补
        }
    }
    // 尾行 间隔为 nInterval
    for (int i = numRows - 1; i < nLen; i += nInterval)
    {
        strRes[nIndex++] = s[i];
    }
    return strRes;
}