给定m个有序（递增）序列， 这些序列中包含的总的元素数目为n, 请将这些数组元素合并为一个有序（递增）序列。算法复杂度O(nm)

解题思路：

先将一个数组（默认第一个数组）拷贝进结果数组，然后将结果数组与待排序数组进行两两合并（相当于两个有序数组的合并）。注意在合并之前，要先将结果数组保存一下。

int * merge(int **values, int *sizes, int m) {
	//先计算大小总的大小
	int count = 0;
	for (int i = 0; i < m; i++)
	{
		count += sizes[i];
	}
	int *merge = new int[count];
	
        //count用来保存已排序元素的个数
	count = sizes[0];
	memcpy(merge, values[0], sizes[0] * 4);
	for (int i = 1; i < m; i++)
	{
		int *temp = new int[count];
		memcpy(temp, merge, count * 4);
		
		int index = 0;
		int a = 0;
		int b = 0;
		while (a < sizes[i] && b < count)
		{
			if (values[i][a] < temp[b])
			{
				merge[index] = values[i][a];
				++a;
				++index;
			}
			else
			{
				merge[index] = temp[b];
				++b;
				++index;
			}
		}
		//将剩余元素拷贝
		if (a == sizes[i])
		{
			while (b < count)
			{
				merge[index] = temp[b];
				++b;
				++index;
			}
		}
		else
		{
			while (a < sizes[i])
			{
				merge[index] = values[i][a];
				++a;
				++index;
			}
		}

		delete[] temp;
                //更新count
		count = index;
	}

	return merge;
}

int main()
{
	int num1[] = { -123,-21,-1,1,3,13,16,19 };
	int num2[] = { 2, 4, 6, 8, 10 ,342,5454,5654,5767};
	int num3[] = { -1, 8, 9, 10, 11 ,34,54,56,76,865};
	int num4[] = { 12, 23, 34, 45, 100,3454,5665,6765,675675 };



	int *values[] = { num1, num2, num3, num4 };
	int sizes[] = { sizeof(num1) / 4, sizeof(num2) / 4 ,sizeof(num3) / 4, sizeof(num4) / 4 };
	int * res = merge(values, sizes,4);


	for (int i = 0; i < sizeof(num1) / 4 + sizeof(num2) / 4 + sizeof(num3) / 4 + sizeof(num4) / 4; ++i)
	{
		cout << res[i] << " ";
	}
	cout << endl;

	system("pause");
	return 0;
}