华为笔试算法题汇总

//////////////////////////////////////////////////////////////////////////华为第一题 19:19-19:36 17分钟 
#include  
#include  
 
using namespace std; 
 
bool g_flag[26]; 
void stringFilter(const char *pInputStr, long lInputLen, char *pOutputStr) 
{ 
 assert(pInputStr != NULL); 
 int i = 0; 
 if (pInputStr == NULL || lInputLen <= 1) 
 { 
  return; 
 } 
 const char *p = pInputStr; 
 while(*p != '\0') 
 { 
  if (g_flag[(*p - 'a')]) 
  { 
   p++; 
  }else{ 
   pOutputStr[i++] = *p; 
   g_flag[*p - 'a'] = 1; 
   p++; 
  } 
 } 
 pOutputStr[i] = '\0'; 
} 
int main() 
{ 
 memset(g_flag,0,sizeof(g_flag)); 
 char input[] = "abacacde"; 
 char *output = new char[strlen(input) + 1]; 
 stringFilter(input,strlen(input),output); 
 cout<

//////////////////////////////////////////////////////////////////////////华为第二题 19:40 - 20:10 中间耽误3分钟 
#include  
#include  
 
using namespace std; 
 
void stringZip(const char *pInputStr, long lInputLen, char *pOutputStr) 
{ 
 const char *p = pInputStr; 
 int num = 1; 
 int i = 0; 
 p++; 
 while(*p != NULL) 
 { 
  while(*p == *(p-1)&& *p != NULL) 
  { 
  num++; 
  p++; 
  } 
  if (num > 1) 
  { 
   int size = 0; 
   int temp = num; 
   while(num)    //计算位数 
   { 
    size++; 
    num /= 10; 
   } 
   num = 1; 
 
   for (int j = size; j > 0; j--) 
   { 
    pOutputStr[i+j-1] = '0'+ temp%10; 
    temp /= 10; 
   } 
   i +=size; 
   pOutputStr[i++] = *(p-1); 
   p++; 
  }else{ 
   pOutputStr[i++] = *(p-1); 
   p++; 
  } 
 } 
 pOutputStr[i] = '\0'; 
} 
 
int main() 
{ 
 char input[] = "cccddecc"; 
 char *output = new char[strlen(input) + 1]; 
 stringZip(input,strlen(input),output); 
 cout<

//////////////////////////////////////////////////////////////////////////华为第三题 20:29 - 20:40 
#include  
 
using namespace std; 
 
void arithmetic(const char *pInputStr, long lInputLen, char *pOutputStr) 
{ 
 const char *input = pInputStr; 
  char *output = pOutputStr; 
 int sum = 0; 
 int operator1 = 0; 
 int operator2 = 0; 
 char *temp = new char[5]; 
 char *ope = temp; 
 while(*input != ' ') //获得操作数1 
 { 
  sum = sum*10 + (*input++ - '0'); 
 } 
 input++; 
 operator1 = sum; 
 sum = 0; 
 
 while(*input != ' ') 
 { 
  *temp++ = *input++; 
 } 
 
 input++; 
 *temp = '\0'; 
 
 if (strlen(ope) > 1 ) 
 { 
  *output++ = '0'; 
  *output = '\0'; 
  return; 
 } 
 
 while(*input != '\0') //获得操作数2 
 { 
  sum = sum*10 + (*input++ - '0'); 
 } 
 operator2 = sum; 
 sum = 0; 
 
 switch (*ope) 
 { 
 case '+':itoa(operator1+operator2,pOutputStr,10); 
  break; 
 case '-':itoa(operator1-operator2,pOutputStr,10); 
 break; 
 default: 
  *output++ = '0'; 
  *output = '\0'; 
  return; 
 } 
} 
 
int main() 
{ 
 char input[] = "4 - 7"; 
 char output[] = " "; 
 arithmetic(input,strlen(input),output); 
 cout<

//华为2014年机试题1：输入1--50个数字，求出最小数和最大数的和 
//输入以逗号隔开 
#include 
#define N 50 
void sort(int a[],int n); 
int main(void) 
{ 
 
 
 char str[100]; 
 int a[N]={0}; 
 gets(str);  //要点1：动态的输入1--50个整数，不能确定个数，只能用字符串输入，然后分离出来 
 int i=0; 
 int j=0; 
 int sign=1; 
 while(str[i]!='\0') 
 { 
  if(str[i]!=',') //输入时要在半角输入 
  { 
 
   if(str[i] == '-') //要点:2：有负整数的输入 
   { 
    // i++; //易错点1 
    sign=-1; 
   } 
   else if(str[i]!='\0') //不用else的话，负号也会减去‘0' 
   { 
    a[j]=a[j]*10 + str[i]-'0'; //要点3：输入的可以是多位数 
 
   } 
  } 
  i++; 
  if(str[i]==',' || str[i]=='\0') //这个判断是在i自加以后 
  { 
    a[j]=a[j]*sign; //易错点2 
    sign=1; ////易错点3 
    j++; //j就是a数组的个数 范围0到j-1 
  } 
 
 
 } 
 
 sort(a,j); 
 printf("Max number + Min number = %d",a[0]+a[j-1]); 
 
 return 0; 
} 
void sort(int a[],int n) //选择排序 
{ 
 int i,j; 
 int k; 
 int temp; 
 for(i=0;ia[j]) 
    k=j; 
  } 
  if(i!=k) 
  { 
   temp = a[k]; 
   a[k] = a[i]; 
   a[i] = temp; 
  } 
 } 
 for(i=0;i