考虑两件事：
1. 考虑当 = key 时该移动 l 还是 r
2. 考虑特殊情况：只有两个数且 a[l] 或 a[r] = key 时 mid 该向下取整还是向上取整来避免死循环

(1) 查找第一个和 key 值相等的元素下标

int binary_search(int a[], int n, int key) {
    int l = 0, r = n - 1;
    while (l < r) {
        int mid = l + r >> 1; // 靠l向后移动所以向下取整
        if (a[mid] >= key) r = mid;
        else l = mid + 1;
    }
    if (a[l] == key) return l;
    return -1;
}

(2) 查找最后一个和 key 值相等的元素下标

int binary_search(int a[], int n, int key) {
    int l = 0, r = n - 1;
    while (l < r) {
        int mid = l + r + 1 >> 1; // 靠r向前移动所以向上取整
        if (a[mid] <= key) l = mid;
        else r = mid - 1;
    }
    if (a[l] == key) return l;
    return -1;
}

(3) 查找第一个大于等于 key 值的元素下标

int binary_search(int a[], int n, int key) {
    int l = 0, r = n - 1;
    while (l < r) {
        int mid = l + r >> 1;
        if (a[mid] >= key) r = mid;
        else l = mid + 1;
    }
    if (a[l] >= key) return l;
    return -1;
}

(4) 查找最后一个小于等于 key 值的元素下标

int binary_search(int a[], int n, int key) {
    int l = 0, r = n - 1;
    while (l < r) {
        int mid = l + r + 1 >> 1;
        if (a[mid] <= key) l = mid;
        else r = mid - 1;
    }
    if (a[l] <= key) return l;
    return -1;
}