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1. Find a single dog
Only two numbers in an array appear once, and all other numbers appear twice.
Write a function to find these two numbers that only appear once.
#include<stdio.h>
//一个数组中只有两个数字是出现一次，其他所有数字都出现了两次。
//编写一个函数找出这两个只出现一次的数字。
int * findonly(int arr[], int n){
int i, j, k;
int br[2];
int x = 0;
for (i = 0; i<n; i++)
{
k = 0;
for (j = 0; j<n; j++)
{
if (arr[j] == arr[i])
{
k++;
}
}
if (k == 1)
{
br[x] = arr[i];
x++;
}
}
return br;
}
void main(){
int ar[] = {
0, 2, 3, 2,4,4, 5, 5, 0, 6 ,7,7};
int n = sizeof(ar) / sizeof(ar[0]);
int *p=findonly(ar, n);
printf("%d %d", *p, *(p + 1));
}
Two, simulate strncat
#include<stdio.h>
#include<string.h>
#include<assert.h>
char *my_strncat(char *strDest, const char *strSrc, size_t count)
{
//参数检查和保护
assert(strDest != NULL && strSrc != NULL);
char *pDest = strDest;
const char *pSrc = strSrc;
//先查找Dest字符串的末尾
while (*pDest != '\0')
pDest++;
//实行连接拷贝
while (count-- > 0)
{
*pDest++ = *pSrc++;
}
//增加结束标记
*pDest = '\0';
return strDest;
}
void main()
{
char str1[20] = "Hello";
char *str2 = "world.";
printf("str1 = %s\n", str1);
char *pret = my_strncat(str1, str2, 3);
printf("str1 = %s\n", str1);
printf("str1 = %s\n", pret);
}
Three, simulate strncpy
#include<stdio.h>
#include<string.h>
#include<assert.h>
//模拟实现strncpy
char * my_strncpy(void *dest, const void *src, size_t count)
{
assert(dest != NULL && src != NULL);
char *pdest = (char *)dest;
const char *psrc = (const char *)src;
while (count-- > 0){
*pdest++ = *psrc++;
}
pdest = '\0';
return dest;
}
void main(){
char *str = "abcdse";
char arr[20] = {
0 };
printf("%s\n", my_strncpy(arr, str,4));
}
Fourth, simulate atoi
#include<stdio.h>
//将字符串转化为一个整数
int myatoi(char *str){
long number = 0;
int flag = 1; //判断符号位
if (NULL == str)
{
printf("str is NULL");
return 0;
}
while (*str == ' ') //空格
{
str++;
}
if (*str == '-') //符号位
{
flag = -1;
str++;
}
while ((*str >= '0') && (*str <= '9'))//转化
{
number = number * 10 + *str - '0';//用第一个符号数字和0-
str++;
}
return flag*number;
}
void main(){
char a[] = "-100";
char b[] = "101";
printf("%d\n", atoi(a));
printf("%d\n", myatoi(b));
}