一道有趣的算法题:仿照Excel的列编号,给定一个数字,输出该列编号字符串

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,一个整数除26然后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,经过晚上接近半个多小时的编写,完成的代码如下:

C++代码如下所示:


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#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 ;
}

JAVA代码如下所示:


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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.