ElementUniqueness问题(EU)

给出一组数给出一组数据,, 判断每个数都是唯一性的或者说判断是否存在重复的.

算法思路很简单, 快速排序 + 遍历判断: Max(O(nlogn) + O(n)) = O(nlogn)算法复杂度

代码实现

bool IsEelementUniqueness(const vector<float>& Elemnts)
{
	vector<float> TempElemnts = Elemnts;
	sort(TempElemnts.begin(), TempElemnts.end());

	for (int index = 0; index < TempElemnts.size() - 1; index++)
	{
		if(TempElemnts[index] == TempElemnts[index + 1])
			return false;
	}

	return true;
}

	vector<float> floatValues = { 4.5f, 2.0f, 3.0f, 2.0f, 1.0f, 1.3f};
	if (IsEelementUniqueness(floatValues))
	{
		printf("Element uniequeness\n");
	}
	else
	{
		printf("not element uniequeness\n");
	}

MinMap和MaxGap问题

MinMap

给出一组数，求相邻两个数的最小间距

求解算法和上面 "EU"问题是同个问题. 相邻两个数最小间距不为0时, 则满足EU，反过来不满足EU.

MaxMap

给出一组数，求相邻两个数的最大间距 

用桶分法:

[1]求出最大和最小值MaxValue和MinValue

[2]n个数，则设计n个Section区间,  SectionLength = (MaxValue - MinValue) / (n - 2)。每个Section存在: 数的数量, 最大值, 最小值

class SectionBucket
{
public:
	float maxValue;
	float minValue;
	int count;

	SectionBucket()
	{
		maxValue = FLT_MIN;
		minValue = FLT_MAX;
		count = 0;
	}
};

[3]遍历所有数, 并且把相应的数用除法划分到相应的区间，并更新该区间的最大和最小值

[4]最后遍历所有的Section, 求取前后两个相邻Section的最大间距.(后Section的最小值减去前Section的最大值)

代码实现

float GetMaxCap(const vector<float>& Elemnts)
{
	float maxValue = FLT_MIN;
	float minValue = FLT_MAX;

	// get max value and min value
	for (int index = 0; index < Elemnts.size(); index++)
	{
		if (Elemnts[index] > maxValue)
		{
			maxValue = Elemnts[index];
		}

		if (Elemnts[index] < minValue)
		{
			minValue = Elemnts[index];
		}
	}

	int n = Elemnts.size();
	vector<SectionBucket> buckets;
	buckets.resize(n);

	float bucketSize = (maxValue - minValue) / (float)(n - 1);

	for (int index = 0; index < n; index++)
	{
		int bucketIndex = int((Elemnts[index] - minValue) / bucketSize) + 1;
		if (bucketIndex == n)
			bucketIndex = n - 1;

		buckets[bucketIndex].count++;

		if (Elemnts[index] > buckets[bucketIndex].maxValue)
		{
			buckets[bucketIndex].maxValue = Elemnts[index];
		}

		if (Elemnts[index] < buckets[bucketIndex].minValue)
		{
			buckets[bucketIndex].minValue = Elemnts[index];
		}
	}


	// bucket index from 1 to n - 1 
	float cap = 0.0f;
	float leftValue = buckets[1].maxValue;
	for (int index = 2; index < n; index++)
	{
		if (buckets[index].count > 0)
		{
			float distance = buckets[index].minValue - leftValue;
			if (distance > cap)
			{
				cap = distance;
			}

			leftValue = buckets[index].maxValue;
		}
	}

	return cap;
}

测试

	vector<float> floatValues = { 4.5f, 2.0f, 3.0f, 2.0f, 1.0f, 1.3f };
	float maxCap = GetMaxCap(floatValues);
	printf("MaxCap = %f\n", maxCap);

算法复杂度

从代码实现很显然MaxCap算法复杂度是O(n)

参考资料

[1]清华计算几何 P61-P65