c语言统计一个数字在排序数组中出现的次数

Posted 2020-09-18 gavanwanggw

// 题目：统计一个数字在排序数组中出现的次数。

//  比如：排序数组｛1。2，3，3，3，3，4。5｝和数字3，因为3出现了4次。因此输出4

有一种最简单的算法，遍历。可是有比它效率更高的

先看遍历：

#include <stdio.h>
#include <assert.h>

int num_time(int *arr, int len, int a)
{
	int i = 0;
	int count = 0;
	assert(arr != NULL);
	for (; i < len; ++i)
	{
		if (arr[i] == a)
			count++;
	}
	return count;
}

int main()
{
	int arr[] = { 1, 2, 3, 3, 3, 3, 4, 5 };
	int len = sizeof(arr) / sizeof(arr[0]);
	printf("%d\n", num_time(arr, len, 3));
	return 0;
}

另一种利用二分查找：

#include <stdio.h>

int GetFirstKey(int arr[], int left, int right, int len, int key)
{
	int mid;
	if (left > right)
	{
		return -1;
	}
	mid = left - (left - right) / 2;
	if (key == arr[mid])
	{
		if ((mid > 0 && arr[mid - 1] != key) || mid == 0)
		{
			return mid;
		}
		else
		{
			right = mid - 1;
		}
	}
	else if (arr[mid] < key)
	{
		left = mid + 1;
	}
	else
	{
		right = mid - 1;
	}
	return GetFirstKey(arr, left, right, len, key);
}

int GetLastKey(int arr[], int left, int right, int len, int key)
{
	int mid;
	if (left > right)
	{
		return -1;
	}
	mid = left - (left - right) / 2;
	if (key == arr[mid])
	{
		if ((mid < len - 1 && arr[mid + 1] != key || mid == len - 1))
		{
			return mid;
		}
		else
		{
			left = mid + 1;
		}
	}
	else if (arr[mid] < key)
	{
		left = mid + 1;
	}
	else
	{
		right = mid - 1;
	}
	return GetLastKey(arr, left, right, len, key);
}

int  main()
{
	int brr[] = { 1, 2, 3, 3, 3, 3, 4, 5};
	int len = sizeof(brr) / sizeof(brr[0]);
	int first = GetFirstKey(brr, 0, len - 1, len - 1, 3);
	int last = GetLastKey(brr, 0, len - 1, len - 1, 3);
	printf("%d\n", last - first + 1);
	return 0;
}