#include <stdio.h>

#define max(a, b) (a < b) ? (b) : (a)
#define min(a, b) (a < b) ? (a) : (b)

int findMaxConsecutiveValues(int a[], int n, int val) {
    int j = 0, cnt = 0;
    for (int i = 0; i < n; i++)
    {
        for (; i < n && val == a[i]; cnt = max(cnt, i - j + 1), i++);
        if (i < n && val != a[i]) j = i + 1;
    }
    return cnt;
}

int findMinConsecutiveValues(int a[], int n, int val) {
    int j = 0, cnt = n, m = 0;  // j慢指针, 连续val的最小个数, m当前遍历到的连续val的个数
    for (int i = 0; i < n; i++) {
        for (; i < n && val == a[i]; i++)  // 统计每一片连续val值的个数
            m++;

        if (i < n && val != a[i]) {        // 连续的val值中断时 
            if (m > 0)                     // m大于0时才能与上一片连续的val值的个数取min
                cnt = min(cnt, m);         // 获取当前最小连续val值的个数
            m = 0;                         // 清零才能正确进行下一片连续val值的统计
            j = i + 1;                     // 慢指针j指向刚统计好的连续val值的下一个数
        }
    }
    return cnt;
}

int main() {
    int a[] = { 0, 1, 1, 1, 1, 0, 1, 1, 2, 2, 2, 1, 1, 2, 2, 11, 0, 2, 2, 2, 2, 2, 1, 1 };
    int n = sizeof(a) / sizeof(a[0]);

    for (int i = 0; i < n; i++)
        printf("%d ", a[i]);
    printf("\n");

    int val = 2;
    printf("序列中连续且最少的%d的%d个数\n", val, findMinConsecutiveValues(a, n, val));

    val = 1;
    printf("序列中连续且最多的%d的%d个数\n", val, findMaxConsecutiveValues(a, n, val));

    return 0;
}