By Long Luo

最近遇到一个算法题：

仿照Excel的列编号，给出一个数字，输出该列编号字符串。

例如：A对应1，Z对应26，AA对应27，AZ对应52 ......

这个题目是一个典型的26进制思路去处理，但是这个题目里面有很多陷阱，在1, 26, 52等特殊情况进行考虑，经过晚上接近1个小时的编写，完成的代码如下：

C++代码如下：

#include <iostream>
#include <string.h>

using namespace std;

//函数itos：正整数到编号转换
//num：输入的正整数，pcout：输出，Max：输出控件最大长度
void itos(int num, char *pcout )
{
     char *res = new char[255];
     int m = 0, n = 0;

     while((num >= 1) && (n < 255))
    {
         m = num % 26;
          if (m != 0)
         {
             res[n] = ‘A‘ + m - 1;
         }
          else
         {
             res[n] = ‘Z‘;
              num--;
         }
         
          num /= 26;
         n++;
    }

     for(m = n; m > 0; m--)
    {
          pcout[n - m] = res[m - 1];
    }

     pcout[n] = ‘\0‘ ;
     delete[] res;

     return;
}

//soti：字符串到数字的转换
int stoi(char *cha)
{
     int m = 0, n = 0, i = 0, val = 0, a = 0;
     char *pc = cha ;

     while(*pc != ‘\0‘ )
    {
          //后移到个位 
         pc++;
         n++;
    }
    
     for(i = 1; i <= n; i++)
    {
          //位循环
         pc--;
         a = i;
         m = 1;
         
          while(a > 1)
         {
              //位权
             m *=26;
             a--;
         }
         
         m *= (*pc - ‘A‘ +1);
         val += m;
    }
    
     return val;
}

int main()
{
     char out[255] = {0};
    printf( "out = %s\n", out);
    itos(32, out);
    printf( "out = %s\n", out);

    getchar();

     return true ;
}

package com.Algorithms.excelrow;

/*
 * @author: Long Luo
 * @Created By Frank Luo @2014.05.01
 */
public class ExcelRow {
	public static void main(String args[]) {

		System.out.println("25=" + int2Str(5) + ",28=" + int2Str(28) + ",123="
				+ int2Str(123));
		System.out.println("C=" + str2Int("C") + ",ZA" + str2Int("ZA")
				+ ",AAF=" + str2Int("AAF"));
	}

	/*
	 * @Description: covert the String to Integer.
	 */
	public static int str2Int(String input) {
		int val = 0;
		int len = input.length();
		int mul = 0;

		for (int i = len - 1, j = 0; i >= 0; i--, j++) {
			mul = 1;

			int temp = input.charAt(i) - ‘A‘ + 1;
			double weiquan = Math.pow(10, j);
			mul = (int) (temp * weiquan);
			val += mul;

			System.out.println("temp=" + temp + ",weiquan=" + weiquan + ",mul="
					+ mul + ",val=" + val);
		}

		return val;
	}

	/*
	 * @Description: covert the Integer to String.
	 */
	public static String int2Str(int rowNum) {
		StringBuffer temp = new StringBuffer(255);
		char ch;

		while (rowNum >= 1) {

			int i = rowNum % 26;
			if (i != 0) {
				ch = (char) (‘A‘ + i - 1);
				temp = temp.append(ch);
			} else {
				ch = ‘Z‘;
				temp = temp.append(ch);
				rowNum--;
			}

			System.out.println("temp=" + temp + ",ch=" + ch + ",rowNum="
					+ rowNum);
			rowNum /= 26;
		}

		return temp.reverse().toString();
	}

}

以上代码均测试通过。

如有不当错误之处，敬请批评指正，如有更好的方法，也请共同探讨, Thx:-)

Long Luo Created at PM22:25 ~ 22:40 @May 02nd, 2014 at Shenzhen, China.