数据结构排序算法---八大排序复杂度及代码实现_编写任意一种排序算法,写明时间复杂度和空间复杂度

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稳定性：相同的数据排序后，相对位置是否发生改变

一、冒泡排序

时间复杂度：O(N^2)
空间复杂度：O(1)
稳定性：稳定

代码实现

void Swap(int* a, int* b)
{
	int tmp;
	tmp = *a;
	*a = *b;
	*b = tmp;
}
void BubbleSort(int* a, int n)
{
	for (int j = 0; j < n; j++)
	{
		int exchange = 0;
		for (size_t i = 1; i < n-j; i++)
			if (a[i - 1] > a[i])
			{
				Swap(&a[i - 1], &a[i]);
				exchange = 1;
			}
		if (exchange == 0)
		{
			break;
		}
	}
}
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二、直接插入排序

时间复杂度：O(N^2)
空间复杂度：O(1)
稳定性：稳定

代码实现

void InsertSort(int* a, int n)
{
	for (int i = 0; i < n; i++)
	{
		int end = i;
		int tmp = a[end + 1];
		while (end>=0)
		{
			if (tmp < a[end])
				a[end + 1] = a[end];
			else
				break;
			--end;
		}
		a[end + 1] = tmp;
	} 
}
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三、希尔排序

时间复杂度：O(N^1.3)
空间复杂度：O(1)
稳定性：不稳定

代码实现

void ShellSort(int* a, int n)
{
	int gap = n;
	while (gap > 1)
	{
		gap = gap / 3 + 1;
		for (int i = 0; i < n - gap; i++)
		{
			int end = i;
			int tmp = a[end + gap];
			while (end >= 0)
			{
				if (tmp < a[end])
				{
					a[end + gap] = a[end];
					end -= gap;
				}
				else
					break;
			}
			a[end + gap] = tmp;
		}
	}
}
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四、选择排序

时间复杂度：O(N^2)
空间复杂度：O(1)
稳定性：不稳定

代码实现

void SelectSort(int* a, int n)
{
	int begin = 0;
	int end = n - 1;
	while (begin < end)
	{
		int max = begin, min = begin;
		for (int i = begin+1; i <= end; i++)
		{
			if (a[i] > a[max])
				max = i;
			if (a[i] < a[min])
				min = i;
		}
		Swap(&a[begin], &a[min]);
		if (max == begin)
			max = min;
		Swap(&a[end], &a[max]);
		--end;
		++begin;
	}
}
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五、堆排序

时间复杂度：O(N*logN)
空间复杂度：O(1)
稳定性：不稳定

代码实现

void AdjustDown(int* a, int n, int parent)
{
	int child = parent * 2 + 1;
	while (child < n)
	{
		if (child + 1 < n && a[child + 1] > a[child])
		{
			++child;
		}

		if (a[child] > a[parent])
		{
			Swap(&a[child], &a[parent]);
			parent = child;
			child = parent * 2 + 1;
		}
		else
		{
			break;
		}
	}
}

void HeapSort(int* a, int n)
{
	for (int i = (n - 1 - 1) / 2; i >= 0; i--)
	{
		AdjustDown(a, n, i);
	}
	int end = n - 1;
	while (end > 0)
	{
		Swap(&a[0], &a[end]);
		AdjustDown(a, end, 0);
		--end;
	}
}
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六、快速排序

时间复杂度：O(N*logN)
空间复杂度：O(logN)
稳定性：不稳定

代码实现

//三数取中
int GetMidi(int* a, int left, int right)
{
	int midi = (left + right) / 2;
	if (a[left] < a[midi])
	{
		if (a[midi] < a[right])
			return midi;
		else if (a[right] < a[left])
			return left;
		else
			return right;
	}
	else
	{
		if (a[left] < a[right])
			return left;
		else if (a[right] < a[midi])
			return midi;
		else
			return right;
	}
}
//hoare
int PartSort1(int* a, int left,int right)
{
	int keyP = left;//GetMidi(a,left,right);
	while (left < right)
	{
		while (a[right] >= a[keyP] && left < right)
			--right;
		while (a[left] <= a[keyP] && left < right)
			++left;
		Swap(&a[left], &a[right]);
	}
	Swap(&a[keyP], &a[left]);
	return left;
}
//挖坑法
int PartSort2(int* a, int left, int right)
{
	int midi = GetMidi(a, left, right);
	Swap(&a[left], &a[midi]);
	int keyP = a[left];
	//int keyP = left;
	int hole = left;
	while (left < right)
	{
		while (a[right] >= keyP && left < right)
			--right;
		a[hole] = a[right];
		hole = right;
		while (a[left] <= keyP && left < right)
			++left;
		a[hole] = a[left];
		hole = left;
	}
	a[hole] = keyP;
	return hole;
}
//前后指针
int PartSort3(int* a, int left, int right)
{
	int prev = left;
	int cur = prev + 1;
	int keyP = left;
	while (cur <= right)
	{
		if (a[cur] < a[keyP] && ++prev != cur)
		{
			Swap(&a[prev], &a[cur]);
		}
		cur++;
	}
	Swap(&a[prev], &a[keyP]);
	return prev;
}

void QuickSort(int* a, int begin, int end)
{
	if (begin >= end)
		return;
	int key = PartSort1(a, begin, end);
	QuickSort(a, begin, key - 1);
	QuickSort(a, key + 1, end);
}

void QuickSort1(int* a, int begin, int end)
{
	if (begin >= end)
		return;
	//小区间优化,小区间不在递归分割排序，降低递归次数
	if ((end - begin + 1) > 10)
	{
		int key = PartSort1(a, begin, end);
		QuickSort1(a, begin, key - 1);
		QuickSort1(a, key + 1, end);
	}
	else
	{
		InsertSort(a + begin, end - begin + 1);
	}
}

void QuickSortNonR(int* a, int begin, int end)//非递归
{
	Stack st;
	StackInit(&st);
	StackPush(&st, end);
	StackPush(&st, begin);
	while (!StackEmpty(&st))
	{
		int left = StackTop(&st);
		StackPop(&st);
		int right = StackTop(&st);
		StackPop(&st);
		int key = PartSort3(a, left, right);
		if (key + 1 < right)
		{
			StackPush(&st, right);
			StackPush(&st, key + 1);
		}
		if (begin < key - 1)
		{
			StackPush(&st, key - 1);
			StackPush(&st, begin);
		}
	}
	StackDestory(&st);
}
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其中包含了，三数取中（基准值），快排的三种实现方法（hoare，挖坑法，前后指针）及非递归方法

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七、归并排序

时间复杂度：O(N*logN)
空间复杂度：O(N)
稳定性：稳定

代码实现

void PartMergeSort(int* a, int* tmp, int begin, int end)
{
	if (end <= begin)
		return;
	int mid = (begin + end) / 2;
	PartMergeSort(a, tmp, begin, mid);
	PartMergeSort(a, tmp, mid + 1, end);
	int begin1 = begin, end1 = mid;
	int begin2 = mid + 1, end2 = end;
	int count = begin;
	while (begin1 <= end1 && begin2 <= end2)
	{
		if (a[begin1] < a[begin2])
			tmp[count++] = a[begin1++];
		else
			tmp[count++] = a[begin2++];
	}
	while (begin1 <= end1)
	{
		tmp[count++] = a[begin1++];
	}
	while (begin2 <= end2)
	{
		tmp[count++] = a[begin2++];
	}
	memcpy(a + begin, tmp + begin, (end - begin + 1) * sizeof(int));
}

void MergeSort(int* a, int n)
{
	int* tmp = malloc(sizeof(int) * n);
	if (tmp == NULL)
	{
		perror("malloc fail");
		exit(-1);
	}
	PartMergeSort(a, tmp, 0, n - 1);
	free(tmp);
}

void MergeSortNonR(int* a, int n)
{
	int* tmp = malloc(sizeof(int) * n);
	if (tmp == NULL)
	{
		perror("malloc fail");
		exit(-1);
	}
	int gap = 1;
	while (gap < n)
	{
		for (int i = 0; i < n; i += 2 * gap)
		{
			int begin1 = i, end1 = i + gap - 1;
			int begin2 = i + gap, end2 = i + 2 * gap - 1;
			int count = i;
			if (begin2 >= n)
				break;
			if (end2 >= n)
				end2 = n - 1;
			while (begin1 <= end1 && begin2 <= end2)
			{
				if (a[begin1] < a[begin2])
					tmp[count++] = a[begin1++];
				else
					tmp[count++] = a[begin2++];
			}
			while (begin1 <= end1)
			{
				tmp[count++] = a[begin1++];
			}
			while (begin2 <= end2)
			{
				tmp[count++] = a[begin2++];
			}
			memcpy(a + i, tmp + i, (end2-i+1) * sizeof(int));
		}	
		gap *= 2;
	}
}
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其中包含了递归及非递归实现方法

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八、计数排序

时间复杂度：O(N+range)
空间复杂度：O(range)

代码实现

void CountSort(int* a, int n)
{
	int min = a[0], max = a[0];
	for (size_t i = 0; i < n; i++)
	{
		if (a[i] < min)
			min = a[i];

		if (a[i] > max)
			max = a[i];
	}

	int range = max - min + 1;
	int* count = (int*)malloc(sizeof(int) * range);
	printf("\nrange:%d\n", range);
	if (count == NULL)
	{
		perror("malloc fail");
		return;
	}
	memset(count, 0, sizeof(int) * range);

	// 统计数据出现次数
	for (int i = 0; i < n; i++)
	{
		count[a[i] - min]++;
	}

	// 排序
	int j = 0;
	for (int i = 0; i < range; i++)
	{
		while (count[i]--)
		{
			a[j++] = i + min;
		}
	}
}
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