华为2道机试题（review）

华为2道机试题：

/**@date:2010/09/14@author:weedge@comment:1. 识别字符串中的整数并转换为数字形式问题描述： 识别输入字符串中所有的整数，统计整数个数并将这些字符串形式的整数转换为数字形式整数。要求实现函数： void take_num(const char *strIn, int *n, unsigned int *outArray)【输入】 strIn： 输入的字符串【输出】 n： 统计识别出来的整数个数 outArray：识别出来的整数值，其中outArray[0]是输入字符串中从左到右第一个整数， outArray[1]是第二个整数，以此类推。数组地址已经分配，可以直接使用【返回】 无注：I、 不考虑字符串中出现的正负号(+, -)，即所有转换结果为非负整数（包括0和正整数）II、 不考虑转换后整数超出范围情况，即测试用例中可能出现的最大整数不会超过unsigned int可处理的范围III、 需要考虑 '0' 开始的数字字符串情况，比如 "00035" ，应转换为整数35； "000" 应转换为整数0；"00.0035" 应转换为整数0和35（忽略小数点：mmm.nnn当成两个数mmm和nnn来识别）IV、 输入字符串不会超过100 Bytes，请不用考虑超长字符串的情况。示例 输入：strIn = "ab00cd+123fght456-25 3.005fgh"输出：n = 6outArray = {0, 123, 456, 25, 3, 5}*/#include <stdio.h>#include <stdlib.h>#include <string.h>void take_num(const char *strIn, int *n, unsigned int *outArray){if(strIn == NULL){printf("this is null string.");}else{int i = 0;int j = 0;int len = strlen(strIn);while(j<=len){if(*strIn >= '0' && *strIn <='9'){const char *begstr = strIn;const char *endstr = strIn;while(*strIn >= '0' && *strIn <='9'){if(*strIn == '0'){begstr = strIn;}endstr = strIn;strIn++;}if(*endstr != '0'){/**(++endstr) = '/0';*/outArray[i] = atoi(begstr);/*atoi() just read back the front number*/i++;}else{outArray[i] = 0;i++;}}strIn++;j++;}*n = i;}}int main(){char *strIn = "ab00cd+123fght456-25 3.005fgh";int n=0;unsigned int outArray[20];take_num(strIn,&n,outArray);printf("n:%d/n",n);for(int i=0; i<n; i++){printf("outArray[%d]=%u/n",i,outArray[i]);}system("pause");return 0;} 

review：

/**@date:2011/09/02 review (not to use the fun atoi())@author:weedge*/#include <stdio.h>#include <stdlib.h>void take_num(const char *strIn, int *n, unsigned int *outArray){if(strIn == NULL){printf("this is null string.");}else{int i = 0;while(*strIn){if(*strIn >= '0' && *strIn <='9'){outArray[i] = 0;while(*strIn >= '0' && *strIn <='9'){outArray[i] = outArray[i]*10 + ((*strIn++) - '0');}i++;}strIn++;}*n = i;}}void take_num1(const char *strIn, int *n, unsigned int *outArray){ //use atoi()if(strIn == NULL){ printf("this is null string."); }else{ int i = 0; while(*strIn){ if(*strIn >= '0' && *strIn <='9'){ const char *begstr = strIn; while(*strIn >= '0' && *strIn <='9'){ strIn++; } outArray[i] = atoi(begstr);/*atoi() just read back the front number*/ i++; } strIn++;} *n = i; } } int main(){char *strIn = "ab00cd+123000fght456-25  3.005fgh";int n=0;unsigned int outArray[20];take_num1(strIn,&n,outArray);printf("n:%d/n",n);for(int i=0; i<n; i++){printf("outArray[%d]=%u/n",i,outArray[i]);}system("pause");return 0;}

/**@date:2010/09/14@author:weedge@comment:问题描述： 在路由器中，一般来说转发模块采用最大前缀匹配原则进行目的端口查找，具体如下：IP地址和子网地址匹配：IP地址和子网地址所带掩码做AND运算后，得到的值与子网地址相同，则该IP地址与该子网匹配。比如：IP地址：192.168.1.100子网：192.168.1.0/255.255.255.0，其中192.168.1.0是子网地址，255.255.255.0是子网掩码。192.168.1.100&255.255.255.0 = 192.168.1.0，则该IP和子网192.168.1.0匹配IP地址：192.168.1.100子网：192.168.1.128/255.255.255.192192.168.1.100&255.255.255.192 = 192.168.1.64，则该IP和子网192.168.1.128不匹配最大前缀匹配：任何一个IPv4地址都可以看作一个32bit的二进制数，比如192.168.1.100可以表示为：11000000.10101000.00000001.01100100，192.168.1.0可以表示为11000000.10101000.00000001.00000000最大前缀匹配要求IP地址同子网地址匹配的基础上，二进制位从左到右完全匹配的位数尽量多（从左到右子网地址最长）。比如：IP地址192.168.1.100，同时匹配子网192.168.1.0/255.255.255.0和子网192.168.1.64/255.255.255.192，但对于子网192.168.1.64/255.255.255.192，匹配位数达到26位，多于子网192.168.1.0/255.255.255.0的24位，因此192.168.1.100最大前缀匹配子网是192.168.1.64/255.255.255.192。 请编程实现上述最大前缀匹配算法。要求实现函数： void max_prefix_match(const char *ip_addr, const char *net_addr_array[], int *n)【输入】ip_addr：IP地址字符串，严格保证是合法IPv4地址形式的字符串 net_addr_array：子网地址列表，每一个字符串代表一个子网，包括子网地址和掩码， 表现形式如上述，子网地址和子网掩码用’/’分开，严格保证是 合法形式的字符串；如果读到空字符串，表示子网地址列表结束【输出】n：最大前缀匹配子网在*net_addr_array[]数组中对应的下标值。如果没有匹配返回-1示例 输入：ip_addr = "192.168.1.100"net_addr_array[] = {"192.168.1.128/255.255.255.192","192.168.1.0/255.255.255.0","192.168.1.64/255.255.255.192","0.0.0.0/0.0.0.0",""}输出：n = 2*/#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX_BUFFER 8#define MAX_NETADDR 100void max_prefix_match(const char *ip_addr, const char *net_addr_array[], int *n){bool matched = false;/*if ip address is matched with subnet, the matching is true.*/if(ip_addr == NULL){printf("ip address is null/n");exit(-1);}/*get the ip address;*/int ip_address[4] = {0,0,0,0};while(*ip_addr != '.') {ip_address[0] = ip_address[0]*10+(*ip_addr-'0');ip_addr++;}ip_addr++;while(*ip_addr != '.') {ip_address[1] = ip_address[1]*10+(*ip_addr-'0');ip_addr++;}ip_addr++;while(*ip_addr != '.') {ip_address[2] = ip_address[2]*10+(*ip_addr-'0');ip_addr++;}ip_addr++;while(*ip_addr != '/0') {ip_address[3] = ip_address[3]*10+(*ip_addr-'0');ip_addr++;}int max_prefix = 0;/*save the max prefix value each ip&subnet match.*/int prefix[MAX_NETADDR];/*save the prefix value each ip&subnet match.*/int i = 0;while(strcmp(net_addr_array[i],"")){int subnet_addr[4];int mask_addr[4];/*copy ip address to compute the binary of ip address*/int copy_ip_addr[4];copy_ip_addr[0] = ip_address[0];copy_ip_addr[1] = ip_address[1];copy_ip_addr[2] = ip_address[2];copy_ip_addr[3] = ip_address[3];/*const char *begstr = net_addr_array[i];const char *endstr = net_addr_array[i];*//*get the subnet address;*/subnet_addr[0] = atoi(net_addr_array[i]);while(*net_addr_array[i] != '.') net_addr_array[i]++;subnet_addr[1] = atoi(++net_addr_array[i]);while(*net_addr_array[i] != '.') net_addr_array[i]++;subnet_addr[2] = atoi(++net_addr_array[i]);while(*net_addr_array[i] != '.') net_addr_array[i]++;subnet_addr[3] = atoi(++net_addr_array[i]);/*get the mask address;*/while(*net_addr_array[i] != '/') net_addr_array[i]++;mask_addr[0] = atoi(++net_addr_array[i]);while(*net_addr_array[i] != '.') net_addr_array[i]++;mask_addr[1] = atoi(++net_addr_array[i]);while(*net_addr_array[i] != '.') net_addr_array[i]++;mask_addr[2] = atoi(++net_addr_array[i]);while(*net_addr_array[i] != '.') net_addr_array[i]++;mask_addr[3] = atoi(++net_addr_array[i]);if(subnet_addr[0] == (ip_address[0]&mask_addr[0]) &&subnet_addr[1] == (ip_address[1]&mask_addr[1]) &&subnet_addr[2] == (ip_address[2]&mask_addr[2]) &&subnet_addr[3] == (ip_address[3]&mask_addr[3])){/*ip address is matched with subnet address.*/matched = true;prefix[i] = 0;int ip_buffer[MAX_BUFFER],subnet_buffer[MAX_BUFFER];int k;for(k=0; k<MAX_BUFFER; k++){ip_buffer[k] = 0;subnet_buffer[k] = 0;}for(int j=0; j<4; j++){k = 0;while(subnet_addr[j]>0){subnet_buffer[k++] = subnet_addr[j]%2;subnet_addr[j] = subnet_addr[j]/2;}k = 0;while(copy_ip_addr[j]>0){ip_buffer[k++] = copy_ip_addr[j]%2;copy_ip_addr[j] = copy_ip_addr[j]/2;}k = 0;while(ip_buffer[k] == subnet_buffer[k] && k<8){prefix[i]++;k++;}}if(max_prefix<prefix[i]){/*get the max prefix and get the suffix of the net_addr_array[].*/max_prefix = prefix[i];/*printf("i = %d/n",i);*/*n = i;}}else{/*ip address is not matched with subnet address.*/prefix[i] = 0;}i++;}/*end while*/if(!matched) printf("there are not the matched net addresses! so max prefix is 0./n");}int main(){/*test*/char *ip_addr = "192.168.1.100";const char *net_addr_array[] = {"192.168.1.128/255.255.255.192","192.168.1.0/255.255.255.0","192.168.1.64/255.255.255.192","0.0.0.0/0.0.0.0",""};int n = 0;/*the suffix of the net_addr_array[].*/max_prefix_match(ip_addr,net_addr_array,&n);printf("n = %d./n",n);system("pause");return 0;}