题目：字符串排序

实现一个函数，输入为一个字符串只包含a-z，返回字符串中出现次数最多的字符及其出现次数，如果有出现同样次数的，则返回在字符串中位置更靠前的

#include <stdio.h>
int get_max_apperance(const char *str, char *most_apperance_char, int *max_times);
int main()
{
   /*  Write C code in this online editor and run it. */
   printf("请输入字符串只包含a-z\n");
	char str[1024]={0};
	char c;
	int times;
	int ret ;
	
	//scanf("%s", &str);
	printf("str=%s\n", str);
	ret = get_max_apperance(str, &c, &times);
	printf("执行结果:%s\n", ret==1?"失败":"成功");
	if(0 == ret){
   		printf("字符串%s中出现次数最多的字符:%c\n出现次数%d\n", str, c, times);
	}
   return 0;
}

/*start time 202301291521*/
/*实现一个函数，输入为一个字符串只包含a-z，返回字符串中出现次数最多的字符及其出现次数，如果有出现同样次数的，则返回在字符串中位置更靠前的*/
/*入参：str 字符串
出参：char 出现次数最多的字符
	 times 出现次数
返回值：0 代表成功
		1 代表失败(比如字符串为空、出现了超出范围a-z的字符)
*/
int get_max_apperance(const char *str, char *most_apperance_char, int *max_times){
	int ret = 0;
	struct time_counter{
		int times;//出现次数
		int first_pos;//第一次出现位置
	};
	struct time_counter counter_arr[26]={0};
	char c;
	int times;
	int i;
	int max_apperance_index;
	if(NULL == str){
		return 1;//字符串为空
	}
	//初始化
	for(i=0;i<26;i++){
		counter_arr[i].times = 0;
		counter_arr[i].first_pos = -1;
	}
	for(i=0;str[i]!='\0';i++){
		c = str[i] - 'a';
		if(c < 0 || c > 25){
			return 1;//超过范围
		}
		counter_arr[c].times++;
		if(-1 == counter_arr[c].first_pos){
			counter_arr[c].first_pos = i;
		}
	}
	max_apperance_index=0;
	for(i=0;i<26;i++){
		if(counter_arr[max_apperance_index].times < counter_arr[i].times){
			max_apperance_index = i;
		}
		if(counter_arr[max_apperance_index].times == counter_arr[i].times && counter_arr[max_apperance_index].first_pos > counter_arr[i].first_pos){
			max_apperance_index = i;
		}
	}
	*most_apperance_char = max_apperance_index + 'a';
	*max_times = counter_arr[max_apperance_index].times;
	return 0;
};
//end coding time 202301291539

//end testing time 202301291607