[翻译转载] Sparse Set 稀疏集

翻译自 Sparse Set - GeeksforGeeks

如果有大量的查询，如何有效地进行以下操作

插入
删减
搜索
清理/删除所有元素

一种解决方案是使用自平衡二进制搜索树，如红黑树、AVL树等。这种解决方案的插入、删除和搜索的时间复杂度为O(Log n)。

我们也可以使用哈希法。在散列法中，前三个操作的时间复杂度是O(1)。但第四个操作的时间复杂度是O(n)。

我们也可以使用 bit-vector (或直接访问表)，但 bit-vector 也需要O(n)时间进行清除。

稀疏集的性能优于所有的BST、哈希和位向量。我们假设给定数据范围（或一个元素的最大值）和可以存储在集合中的最大元素数。我们的想法是维护两个数组：sparse[]和 dense[]。

dense[]   ==> 储存实际的元素
sparse[]  ==> 这就像我们使用元素作为索引的位向量。这里的值不是二进制的，而是密集数组的索引
maxVal    ==> 这个集合可以存储的最大值。sparse[]的大小等于maxVal + 1
capacity  ==> 集的容量。sparse的大小等于capacity。
n         ==> 当前Set中元素的数量

/* A C program to implement Sparse Set and its operations */
#include <algorithm>
#include <iostream>using namespace std;// A structure to hold the three parameters required to
// represent a sparse set.
class SSet
{int *sparse;   // To store indexes of actual elementsint *dense;       // To store actual set elementsint n;           // Current number of elementsint capacity;  // Capacity of set or size of dense[]int maxValue;  // Maximum value in set or size of sparse[]public:// ConstructorSSet(int maxV, int cap){sparse = new int[maxV + 1];dense = new int[cap];capacity = cap;maxValue = maxV;n = 0;     // No elements initially}// Destructor~SSet(){delete[] sparse;delete[] dense;}// If element is present, returns index of// element in dense[]. Else returns -1.int search(int x);// Inserts a new element into setvoid insert(int x);// Deletes an elementvoid deletion(int x);// Prints contents of setvoid print();// Removes all elements from setvoid clear() { n = 0; }// Finds intersection of this set with s// and returns pointer to result.SSet* intersection(SSet &s);// A function to find union of two sets// Time Complexity-O(n1+n2)SSet *setUnion(SSet &s);
};// If x is present in set, then returns index
// of it in dense[], else returns -1.
int SSet::search(int x)
{// Searched element must be in rangeif (x > maxValue)return -1;// The first condition verifies that 'x' is within 'n' in this set and the second// condition tells us that it is present in the data structure.if (sparse[x] < n && dense[sparse[x]] == x)return (sparse[x]);// Not foundreturn -1;
}// Inserts a new element into set
void SSet::insert(int x)
{// Corner cases, x must not be out of// range, dense[] should not be full and// x should not already be presentif (x > maxValue)return;if (n >= capacity)return;if (search(x) != -1)return;// Inserting into array-dense[] at index 'n'.dense[n] = x;// Mapping it to sparse[] array.sparse[x] = n;// Increment count of elements in setn++;
}// A function that deletes 'x' if present in this data
// structure, else it does nothing (just returns).
// By deleting 'x', we unset 'x' from this set.
void SSet::deletion(int x)
{// If x is not presentif (search(x) == -1)return;int temp = dense[n-1]; // Take an element from enddense[sparse[x]] = temp; // Overwrite.sparse[temp] = sparse[x]; // Overwrite.// Since one element has been deleted, we// decrement 'n' by 1.n--;
}// prints contents of set which are also content
// of dense[]
void SSet::print()
{for (int i=0; i<n; i++)printf("%d ", dense[i]);printf("\n");
}// A function to find intersection of two sets
SSet* SSet::intersection(SSet &s)
{// Capacity and max value of result setint iCap = min(n, s.n);int iMaxVal = max(s.maxValue, maxValue);// Create result setSSet *result = new SSet(iMaxVal, iCap);// Find the smaller of two sets// If this set is smallerif (n < s.n){// Search every element of this set in 's'.// If found, add it to resultfor (int i = 0; i < n; i++)if (s.search(dense[i]) != -1)result->insert(dense[i]);}else{// Search every element of 's' in this set.// If found, add it to resultfor (int i = 0; i < s.n; i++)if (search(s.dense[i]) != -1)result->insert(s.dense[i]);}return result;
}// A function to find union of two sets
// Time Complexity-O(n1+n2)
SSet* SSet::setUnion(SSet &s)
{// Find capacity and maximum value for result// set.int uCap = s.n + n;int uMaxVal = max(s.maxValue, maxValue);// Create result setSSet *result = new SSet(uMaxVal, uCap);// Traverse the first set and insert all// elements of it in result.for (int i = 0; i < n; i++)result->insert(dense[i]);// Traverse the second set and insert all// elements of it in result (Note that sparse// set doesn't insert an entry if it is already// present)for (int i = 0; i < s.n; i++)result->insert(s.dense[i]);return result;
}// Driver program
int main()
{// Create a set set1 with capacity 5 and max// value 100SSet s1(100, 5);// // Insert elements into the set set1s1.insert(5);s1.insert(3);s1.insert(9);s1.insert(10);// Printing the elements in the data structure.printf("The elements in set1 are\n");s1.print();int index = s1.search(3);// 'index' variable stores the index of the number to// be searched.if (index != -1) // 3 existsprintf("\n3 is found at index %d in set1\n",index);else         // 3 doesn't existprintf("\n3 doesn't exists in set1\n");// Delete 9 and print set1s1.deletion(9);s1.print();// Create a set with capacity 6 and max value// 1000SSet s2(1000, 6);// Insert elements into the sets2.insert(4);s2.insert(3);s2.insert(7);s2.insert(200);// Printing set 2.printf("\nThe elements in set2 are\n");s2.print();// Printing the intersection of the two setsSSet *intersect = s2.intersection(s1);printf("\nIntersection of set1 and set2\n");intersect->print();// Printing the union of the two setsSSet *unionset = s1.setUnion(s2);printf("\nUnion of set1 and set2\n");unionset->print();return 0;
}

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