字符串分割与比对问题

cjavacjava123 2011-06-28 09:56:45
因为要做个code128条码识别的程序，

这个是条码库，
char *c128[] = {
"11011001100","11001101100","11001100110","10010011000","10010001100",
"10001001100","10011001000","10011000100","10001100100","11001001000",
"11001000100","11000100100","10110011100","10011011100","10011001110",
"10111001100","10011101100","10011100110","11001110010","11001011100",
"11001001110","11011100100","11001110100","11101101110","11101001100",
"11100101100","11100100110","11101100100","11100110100","11100110010",
"11011011000","11011000110","11000110110","10100011000","10001011000",
"10001000110","10110001000","10001101000","10001100010","11010001000",
"11000101000","11000100010","10110111000","10110001110","10001101110",
"10111011000","10111000110","10001110110","11101110110","11010001110",
"11000101110","11011101000","11011100010","11011101110","11101011000",
"11101000110","11100010110","11101101000","11101100010","11100011010",
"11101111010","11001000010","11110001010","10100110000","10100001100",
"10010110000","10010000110","10000101100","10000100110","10110010000",
"10110000100","10011010000","10011000010","10000110100","10000110010",
"11000010010","11001010000","11110111010","11000010100","10001111010",
"10100111100","10010111100","10010011110","10111100100","10011110100",
"10011110010","11110100100","11110010100","11110010010","11011011110",
"11011110110","11110110110","10101111000","10100011110","10001011110",
"10111101000","10111100010","11110101000","11110100010","10111011110",
"10111101110","11101011110","11110101110","10111011110","10111101110",
"11101011110","11010000100","11010010000","11010011100","1100011101011" };

我输入1个长字符串：1101001000010011100110110011100101100011101011，如何比对前11位的字符串str1“11010010000”（code128表示B类码，11位） str2 "10011100110"（表示数字1，11位）str3“11001110010”（校验码，11位）str4“1100011101011”（终止码，13位）

如何写个算法，把长字符串拆为4个字符串（上例），并在条码库中找到与str1，str2，str3，str4相同的字符串？？
非常感谢！
...全文
98 7 打赏 收藏 转发到动态 举报
写回复
用AI写文章
7 条回复
切换为时间正序
请发表友善的回复…
发表回复
赵4老师 2011-06-29
  • 打赏
  • 举报
 回复
小于1000次循环偶从来不考虑效率。
lxyppc 2011-06-28
  • 打赏
  • 举报
 回复
用字符串比较的方式效率太低，可以先把字符串按位转换成数字，在表中查找字符串对应的数字


#include <iostream>

#include <bitset>

#include <map>

#include "stdlib.h"

#include "stdio.h"

#include "string.h"

#include <string>

using namespace std;



const char* c128[] = {

"11011001100","11001101100","11001100110","10010011000","10010001100",

"10001001100","10011001000","10011000100","10001100100","11001001000",

"11001000100","11000100100","10110011100","10011011100","10011001110",

"10111001100","10011101100","10011100110","11001110010","11001011100",

"11001001110","11011100100","11001110100","11101101110","11101001100",

"11100101100","11100100110","11101100100","11100110100","11100110010",

"11011011000","11011000110","11000110110","10100011000","10001011000",

"10001000110","10110001000","10001101000","10001100010","11010001000",

"11000101000","11000100010","10110111000","10110001110","10001101110",

"10111011000","10111000110","10001110110","11101110110","11010001110",

"11000101110","11011101000","11011100010","11011101110","11101011000",

"11101000110","11100010110","11101101000","11101100010","11100011010",

"11101111010","11001000010","11110001010","10100110000","10100001100",

"10010110000","10010000110","10000101100","10000100110","10110010000",

"10110000100","10011010000","10011000010","10000110100","10000110010",

"11000010010","11001010000","11110111010","11000010100","10001111010",

"10100111100","10010111100","10010011110","10111100100","10011110100",

"10011110010","11110100100","11110010100","11110010010","11011011110",

"11011110110","11110110110","10101111000","10100011110","10001011110",

"10111101000","10111100010","11110101000","11110100010","10111011110",

"10111101110","11101011110","11110101110","10111011110","10111101110",

"11101011110","11010000100","11010010000","11010011100","1100011101011" };



#define ArrSize(x)  (sizeof(x)/sizeof(x[0]))



int main()

{

    const char* str = "1101001000010011100110110011100101100011101011";

    map<unsigned long, int>  c123_map;

    // 将条码表存在map中

    for(int i=0;i<ArrSize(c128);i++){

        c123_map[bitset<16>(string(c128[i])).to_ulong()] = i;

    }

    // 根据输入字符串在条码表中查找相应的条码

    while(strlen(str)>21){

        string str1(str,11);

        str+=11;

        // 打印出当前匹配的条码在条码表中的位置

        cout<<str1<<" -> "<<c123_map[bitset<16>(str1).to_ulong()]<<endl;

    }

    // 最后剩余的就是终止码

    cout<<str<<" -> "<<c123_map[bitset<16>(string(str)).to_ulong()]<<endl;

    return 0;

｝



输出结果：

11010010000 -> 107

10011100110 -> 17

11001110010 -> 18

1100011101011 -> 109
赵4老师 2011-06-28
  • 打赏
  • 举报
 回复 
char *c128[110] = {

"11011001100","11001101100","11001100110","10010011000","10010001100",

"10001001100","10011001000","10011000100","10001100100","11001001000",

"11001000100","11000100100","10110011100","10011011100","10011001110",

"10111001100","10011101100","10011100110","11001110010","11001011100",

"11001001110","11011100100","11001110100","11101101110","11101001100",

"11100101100","11100100110","11101100100","11100110100","11100110010",

"11011011000","11011000110","11000110110","10100011000","10001011000",

"10001000110","10110001000","10001101000","10001100010","11010001000",

"11000101000","11000100010","10110111000","10110001110","10001101110",

"10111011000","10111000110","10001110110","11101110110","11010001110",

"11000101110","11011101000","11011100010","11011101110","11101011000",

"11101000110","11100010110","11101101000","11101100010","11100011010",

"11101111010","11001000010","11110001010","10100110000","10100001100",

"10010110000","10010000110","10000101100","10000100110","10110010000",

"10110000100","10011010000","10011000010","10000110100","10000110010",

"11000010010","11001010000","11110111010","11000010100","10001111010",

"10100111100","10010111100","10010011110","10111100100","10011110100",

"10011110010","11110100100","11110010100","11110010010","11011011110",

"11011110110","11110110110","10101111000","10100011110","10001011110",

"10111101000","10111100010","11110101000","11110100010","10111011110",

"10111101110","11101011110","11110101110","10111011110","10111101110",

"11101011110","11010000100","11010010000","11010011100","1100011101011" };

char s[47]="1101001000010011100110110011100101100011101011"

char str1[12],str2[12],str3[12],str4[14];

int i,k1,k2,k3,k4;

strncpy(str1,s   ,11);str1[11]=0;

strncpy(str2,s+11,11);str2[11]=0;

strncpy(str3,s+22,11);str3[11]=0;

strncpy(str4,s+33,13);str4[13]=0;

for (i=0;i<110;i++) if (0==strcmp(str1,c128[i])) break; k1=i;

for (i=0;i<110;i++) if (0==strcmp(str2,c128[i])) break; k2=i;

for (i=0;i<110;i++) if (0==strcmp(str3,c128[i])) break; k3=i;

for (i=0;i<110;i++) if (0==strcmp(str4,c128[i])) break; k4=i;
就想叫yoko 2011-06-28
  • 打赏
  • 举报
 回复
C语言 用 sscanf 分割 后 写个for循环 strcmp，
第二步还可以优化
AnYidan 2011-06-28
  • 打赏
  • 举报
 回复
例子网上应该有很多， 1楼用的是c++,如果不熟悉，看看<string.h>的库函数

你的库实际上是个字符指针数组，
cjavacjava123 2011-06-28
  • 打赏
  • 举报
 回复
这是什么?能不能给点例子,没学过
至善者善之敌 2011-06-28
  • 打赏
  • 举报
 回复
这个你用string 还是CSTIRNG 中的方法都可以实现
没什么难的。再有查找比对的话用4次遍历就可以了
java当中截取字符串和字符串分割的使用
当我想要截取字符串某个字符前面的，或者后面的我们可以采用以下方案，第二种方案也可以采取。 public static void main(String[] args){ String N = "01|大汽车"; String L=""; String R=""; int k= N.length()...
C++_字符串拆分效率比对记录
字符串拆分效率比对记录 目录 字符串拆分效率比对记录 概述 拆分函数 使用regex 使用C函数strtok 使用stringstream 使用string库函数 结果 测试程序 测试结果 概述 自定义字符串拆分函数，分别处理同样...
python 字符串分割_python拆分字符串到列表
多段拆分 使用参数作为分割线将原始字符串拆分成若干个字符串并组织成列表返回 str:比对字符串 包含多个字符串的列表对象 str1 = hello pythonstr1.split...左侧查询3段拆分 从字符串左侧查找到参数后，将参数左侧...
中文字符串分割存入数组 中文字符串顺序匹配（字符串整体相似度） php<二>
将一个中文字符串分割存入数组（ps：公司领导教给我的，在这里拿出来分享了），再进行顺序匹配货的相似度/** * 分割字符串到数组 * @param $string 字符串 * @param int $len 几个组合为一段，默认1 * @return ...
JAVA 字符串三个常用操作（查找子串、截取字符串、分割字符串）
public int indexOf(int ch ) ：返回指定字符在字符串中第一次出现处的索引，如果此字符串中没有这样的字符，则返回 -1。 或 public int indexOf(int ch, int fromIndex) ：返回从 fromIndex 位置开始查找指定字符在...
C语言

70,037

社区成员

243,246

社区内容

发帖
与我相关
我的任务
社区描述
C语言相关问题讨论
社区管理员
  • C语言
  • 花神庙码农
  • 架构师李肯
加入社区
  • 近7日
  • 近30日
  • 至今

加载中

查看更多榜单
社区公告
暂无公告

试试用AI创作助手写篇文章吧

+ 用AI写文章