分治策略与递归: link.

寻找最小差值

通过划分去寻找最小差值，首先将数组划分，将其划分至最小的时候进行差值比较，接着随着递归并不断进行差值比较，从而得到最小差值

int Partition(int* br, int left, int right) //划分函数
{
	int tmp = br[left];
	while (left < right)
	{
		while (left < right && br[right] > tmp)
		{
			--right;
		}
		if (left < right)
		{
			br[left] = br[right];
		}
		while (left < right && br[left] <= tmp)
		{
			++left;
		}
		if (left < right)
		{
			br[right] = br[left];
		}
	}
	br[left] = tmp;
	return left;
}
int FindK(int* br, int left, int right, int k)
{
	if (left == right && k == 1)
	{
		return br[left];
	}
	int pos = Partition(br, left, right);//划分函数
	//left right是物理下标 k是逻辑下标
	int j = pos - left + 1;//该物理位置属于第几小
	if (k <= j)
	{
		return FindK(br, left, pos, k);
	}
	else
	{
		return FindK(br, pos + 1, right, k - j);
	}
}
int MaxS1(const int* br, int left, int right)
{
	return br[right];
}
int MinS2(const int* br, int left, int right)
{
	int mins = br[left];
	for (int i = left + 1; i <= right;i++)
	{
		if (mins > br[i])
		{
			mins = br[i];
		}
	}
	return mins;
}
int Min(int a, int b)
{
	return a < b ? a : b;
}
int Min(int a, int b, int c)
{
	return Min(a, Min(b, c));
}
int Cpair(int* br, int left, int right)
{
	if (right - left <= 0) return INT_MAX;
	int mid = (right - left + 1) / 2;
	FindK(br, left, right, mid); //找到正确中间值位置
	int pos = left + mid - 1;
	int d1 = Cpair(br, left, pos); //左划分
	int d2 = Cpair(br, pos + 1, right); //又划分

	int maxs = MaxS1(br, left, pos);
	int mins = MinS2(br, pos + 1, right);

	return Min(d1, d2, mins - maxs);//左集合差值，右集合差值，两集合差值
}
int Cpair_Ar(int* br, int n)
{
	if (br == NULL || n < 2) return INT_MAX;//无穷大
	else return Cpair(br, 0, n - 1);
}
int main()
{
	int ar[] = { 12,34,78,90,23,45,56,89,100,67 };
	int n = sizeof(ar) / sizeof(ar[0]);
	int dist = Cpair_Ar(ar, n);

	cout << dist << endl;
	return 0;
}

归并排序的非递归实现

void Print_Ar(int* br, int n)
{
	for (int i = 0; i < n; i++)
	{
		cout << br[i] << " ";
	}
	cout << endl;
}
void Merge(int* src, int* dest, int left, int m, int right)
{
	int i = left, j = m + 1;
	int k = left;
	while (i <= m && j <= right)
	{
		dest[k++] = src[i] <= src[j] ? src[i++] : src[j++];	
	}
	while (i <= m)//j->right 结束
	{
		dest[k++] = src[i++];
	}
	while (j <= right)//或者 i->m 结束
	{
		dest[k++] = src[j++];
	}
}
void MergePass(int* src, int* dest, int n, int s)
{
	int i = 0;
	for (i = 0; i + 2 * s - 1<= n - 1; i += s * 2) //保证可以做归并 i+2*s<=n 可以再次进行一次归并
	{                 //初始 左块的末尾  右块末尾  
		Merge(src, dest, i, i + s - 1, i + 2 * s - 1);
	}
	if (n - 1 > i + s - 1) //加一次s不超过总长
	{
		Merge(src, dest, i, i + s - 1, n - 1);//够两个块归并
	}
	else
	{ 
		for (int j = i; j < n; j++)
		{
			dest[j] = src[j]; //n<i+s 不足归并一个块大小，直接进行拷贝
		}
	}
}
void MergeSort(int* br, int n)
{
	if (br == NULL || n < 2) return;
	int* tmp = new int[n];
	int s = 1; //块的大小
	while (s < n)
	{
		MergePass(br, tmp, n, s);
		s += s;
		MergePass(tmp, br, n, s);
		s += s;
	}
	delete[]tmp;
}

int main()
{
	int ar[] = { 12,34,78,90,23,45,56,89,100,67 };
	int n = sizeof(ar) / sizeof(ar[0]);
	Print_Ar(ar, n);
	MergeSort(ar, n);
	Print_Ar(ar, n);

	return 0;
}

全排列问题

void Perm(int* br, int i, int m)
{
	if (i == m)
	{
		for (int j = 0; j <= m; j++)
		{
			cout << br[j] << " ";
		}
		cout << endl;
	}
	else
	{
		for (int j = i; j <= m; j++)
		{
			std::swap(br[i], br[j]);//第一次交换固定
			Perm(br, i + 1, m);
			std::swap(br[i], br[j]);//第二次交换归位
		}
	}
}
int main()
{
	int ar[] = { 1,2,3 };
	int n = sizeof(ar) / sizeof(ar[0]);
	Perm(ar, 0, n - 1);

}

对于分治策略问题，仅仅需要考虑一层的策略，并将其进行分治划分，将大量问题划分为小的问题，其主要关注的就是如何划分以及如何截至

求集合子集

void Print(int *ar,int* br, int i, int n)
{
	if (i > n)
	{
		for (int j = 0; j <= n; j++)
		{
			if (br[j] == 1)
			{
				cout << ar[j] << " ";
			}
		}
		cout << endl;
	}
	else
	{
		br[i] = 1;
		Print(ar, br, i + 1, n);
		br[i] = 0;
		Print(ar, br, i + 1, n);
	}
}
int main()
{
	int ar[] = { 1,2,3 };
	int br[] = { 0,0,0 };
	int n = sizeof(ar) / sizeof(ar[0]);
	Print(ar, br, 0, n - 1);
}

根据深度遍历算法，通过递归来求集合子集，以层次观点去看一层的关系