字符串查找--B中是否有元素不在A中



#include <stdio.h>
int main(int argc, char const *argv[])
{
	char str[26]="AFDKJASD";
	char str2[26]="AAAAZ";
	find_not_include_string(str,str2);
	return 0;
}

void find_not_include_string(char *str,char *sub_str)
{

	int  father_str[26]={0};
	int   sub_str[26]={0};

	if (*str==NULL||*sub_str==NULL)
	{
		return;
	}

	for (int i = 0; i < strlen(str); ++i)
	{
		father_str[str[i]-'A']++;
	}
	for (int i = 0; i < strlen(sub_str); ++i)
	{
		sub_str[sub_str[i]-'A']++;
	}
	for (int i = 0; i < 26; ++i)
	{
		if(father_str[i]==0&&sub_str[i] != 0)
		{

			printf("%c is not in str!
", i+'A');
			break;
		}
	}

	printf("%s
","str2 is include by str!" );
}
原文地址:https://www.cnblogs.com/snake-hand/p/3162949.html