Question

所以我终于认为我已经完成了一个有效的跳过列表并且认为我应该检查我的工作并研究搜索功能的计时。我找到了一个如何使用clock()的教程，并按照以下方式实现：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include "skiplist.h"

int main() {
    // initialize
    SkipList *list = initSkipList();
    // numbers to search for, array to store time
    int n[] = {1, 10, 50, 100, 1000, 5000, 10000, 25000, 50000, 100000, 200000};
    double time[11];

    // insert
    for (int i = 1; i <= 200000; i++)
        insertElement(list, i);

    // search, time, print
    for (int i = 0; i < 11; i++) {
        clock_t t;
        t = clock();
        findElement(list, n[i]);
        t = clock() - t;
        double time_taken = ((double)t)/CLOCKS_PER_SEC; // in seconds
        ti[i] = time_taken;
        printf("fun() took %f seconds to execute \n", time_taken);
    }
    return 0;
}

我想通过这样做并绘制N对时间我会得到一个看似对数的图形，但我的图形看起来是线性的：

我是否应该对如何计算我的功能以尝试测试时间复杂度有误解，或者我的跳过列表是不是应该搜索它？这是我的整个跳过列表实现，以防万一，但搜索功能称为findElement()：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

#ifndef SKIPLIST_H_
#define SKIPLIST_H_


// number of lists
#define MAXLEVEL 5 

// node with pointers
typedef struct Node {
    int data;
    struct Node *next[1]; // or C99 using flexible array members: *next[]
} Node;


// skiplist
typedef struct SkipList {
    Node *header;
    int level;
    int count;
} SkipList;


// initialize skip list
SkipList* initSkipList() {  
    int i;
    SkipList *list = calloc(1, sizeof(SkipList));
    if (!list) {
        printf("Memory Error\n");
        exit(1);
    }
    //list->level = 0;  
    if ((list->header = calloc(1, sizeof(Node) + MAXLEVEL*sizeof(Node*))) == 0) {
        printf("Memory Error\n");
        exit(1);
    }   
    for (i = 0; i <= MAXLEVEL; i++)
        list->header->next[i] = list->header;   // or = list->header?
    printf("Skip list initialized.\n");
    //srand(time(NULL));
    return list;
}

// insert into skip list, return pointer to node
Node* insertElement(SkipList *list,int data) {
    int i, newLevel;
    Node* temp = list->header;
    Node *update[MAXLEVEL+1];

    // find where data belongs
    for (i = list->level; i >= 0; i--) {
        while(temp->next[i] != list->header && temp->next[i]->data < data)
            temp = temp->next[i];
        update[i] = temp;
    }
    temp = temp->next[0];
    // if element already exists, just return it (no duplicates)
    if (temp != list->header && temp->data == data)
        return temp;

    // determine level (coin flips till failure or max level reached)
    for (newLevel = 0; (rand() < RAND_MAX/2) && (newLevel < MAXLEVEL); newLevel++); // Pr(4) == Pr(5)??
    if (newLevel > list->level) {
        for (i = list->level + 1; i <= newLevel; i++)
            update[i] =  list->header;
        list->level = newLevel;
    }

    // make new  node
    if ((temp = calloc(1, sizeof(Node) + newLevel*sizeof(Node*))) == 0) {
        printf("Memory Error\n");
        exit(1);
    }
    temp->data = data;
    list->count++;
    // update next links
    for (i = 0; i <= newLevel; i++) {
        temp->next[i] = update[i]->next[i];
        update[i]->next[i] = temp;
    }

    //printf("Element %d inserted into list. (level %d)\n", data, newLevel);
    return temp;
}


// delete node containing data
void deleteElement(SkipList *list, int data) {
    int i;
    Node *update[MAXLEVEL+1], *temp;
    temp = list->header;
    for (i = list->level; i >= 0; i--) {
        while (temp->next[i] != list->header && temp->next[i]->data < data)
            temp = temp->next[i];
        update[i] = temp;
    }
    // move to (possible) node to delete
    temp = temp->next[0];
    // if doesn't exist
    if (temp == list->header || temp->data != data) {
        printf("Element %d doesn't exist.\n", data);    
        return;
    }
    // adjust next pointers
    for (i = 0; i <= list->level; i++) {
        if (update[i]->next[i] != temp) break;
        update[i]->next[i] = temp->next[i];
    }
    free (temp);
    printf("Element %d deleted from list.\n", data);
    list->count--;
    // adjust header level
    while ((list->level > 0) && (list->header->next[list->level] == list->header)) // double check
        list->level--;
}


// find node containing data
Node* findElement(SkipList *list, int data){
    int i;
    Node *temp = list->header;
    for (i = list->level; i >= 0; i--) {
        while (temp->next[i] != list->header && temp->next[i]->data < data)
            temp = temp->next[i];
    }
    temp = temp->next[0];
    if (temp != list->header && temp->data == data) {
        printf("Element %d found and returned.\n", data);
        return (temp);
    }
    printf("Element %d not found.\n", data);
    return 0;
}


/* Dynamic array for use in printSkipList() function */
typedef struct {
    int *array;
    size_t used;
    size_t size;
} Array;
void initArray(Array *a, size_t initialSize) {
    a->array = malloc(initialSize * sizeof(int));
    a->used = 0;
    a->size = initialSize;
}
void insertArray(Array *a, int element) {
    if (a->used == a->size) {
        a->size *= 2;
        a->array = realloc(a->array, a->size * sizeof(int));
    }
    a->array[a->used++] = element;
}
void freeArray(Array *a) {
    free(a->array);
    a->array = NULL;
    a->used = a->size = 0;
}


// print skip-list info and representational figure
void printSkipList(SkipList *list) {
    int i, j, k, pos = 0, prevPos = 0, diff, numDigits;
    Node* temp = list->header;
    Array a;

    // fill dynamic array with level 0 elements
    initArray(&a, 10);
    while (temp->next[0] != list->header) {
        temp = temp->next[0];
        insertArray(&a, temp->data);
    }
    temp = list->header;
    // print number of levels
    printf("\nNumber of levels in skip-list: %d\n", list->level + 1);
    printf("Number of elements in skip-list: %d\n", list->count);
    printf("Skip-list figure: \n");
    // print data
    for (i = list->level; i >= 0; i--) {
        pos = 0, prevPos = 0;
        while (temp->next[i] != list->header) {
        numDigits = 0;      
            temp = temp->next[i];
            while (temp->data != a.array[pos]) {
                numDigits += floor (log10 (abs (a.array[pos]))) + 1;
                pos++;
            }
            pos++;
            diff = (pos - prevPos) - 1; 
            if (diff >= 1) {
                for (j = 0; j < (4*diff)+numDigits; j++) 
                        printf(" ");    
            }           
            printf("%d -> ", temp->data);
            prevPos = pos;
        }
        temp = list->header;
        printf("\n");       
    }
    printf("\n\n");
}

#endif // SKIPLIST_H_

非常感谢任何建议，谢谢。

Answer 1

你的MAXLEVEL太小了。我认为原始文件的级别高达lg（n），即列表大小的日志基数2。

来自Puch 1990年的原始跳过表：

确定MaxLevel

由于我们可以安全地将水平限制在L（n），我们应该   选择MaxLevel = L（N）（其中N是数字的上限   跳过列表中的元素）。如果p = l / 2，则使用MaxLevel = 16   适用于包含最多2 ^ 16个元素的数据结构

通常，如果p为1 / X，则使用列表大小的log base X.

MAXLEVEL = 5，我得到的结果与你看到的相同。

evaitl@bb ~/se $ ./foo 
Skip list initialized.
Element 1 found and returned.
fun() took 0.000014 seconds to execute 
Element 10 found and returned.
fun() took 0.000002 seconds to execute 
Element 50 found and returned.
fun() took 0.000002 seconds to execute 
Element 100 found and returned.
fun() took 0.000002 seconds to execute 
Element 1000 found and returned.
fun() took 0.000003 seconds to execute 
Element 5000 found and returned.
fun() took 0.000004 seconds to execute 
Element 10000 found and returned.
fun() took 0.000006 seconds to execute 
Element 25000 found and returned.
fun() took 0.000011 seconds to execute 
Element 50000 found and returned.
fun() took 0.000021 seconds to execute 
Element 100000 found and returned.
fun() took 0.000044 seconds to execute 
Element 200000 found and returned.
fun() took 0.000087 seconds to execute

将MAXLEVEL提高到20，我得到：

evaitl@bb ~/se $ ./foo 
Skip list initialized.
Element 1 found and returned.
fun() took 0.000016 seconds to execute 
Element 10 found and returned.
fun() took 0.000003 seconds to execute 
Element 50 found and returned.
fun() took 0.000003 seconds to execute 
Element 100 found and returned.
fun() took 0.000002 seconds to execute 
Element 1000 found and returned.
fun() took 0.000002 seconds to execute 
Element 5000 found and returned.
fun() took 0.000003 seconds to execute 
Element 10000 found and returned.
fun() took 0.000004 seconds to execute 
Element 25000 found and returned.
fun() took 0.000003 seconds to execute 
Element 50000 found and returned.
fun() took 0.000004 seconds to execute 
Element 100000 found and returned.
fun() took 0.000003 seconds to execute 
Element 200000 found and returned.
fun() took 0.000004 seconds to execute

在您的n []：

中添加了400000和800000

evaitl@bb ~/se $ ./foo 
Skip list initialized.
Element 1 found and returned.
fun() took 0.000016 seconds to execute 
Element 10 found and returned.
fun() took 0.000001 seconds to execute 
Element 50 found and returned.
fun() took 0.000001 seconds to execute 
Element 100 found and returned.
fun() took 0.000002 seconds to execute 
Element 1000 found and returned.
fun() took 0.000002 seconds to execute 
Element 5000 found and returned.
fun() took 0.000002 seconds to execute 
Element 10000 found and returned.
fun() took 0.000002 seconds to execute 
Element 25000 found and returned.
fun() took 0.000004 seconds to execute 
Element 50000 found and returned.
fun() took 0.000003 seconds to execute 
Element 100000 found and returned.
fun() took 0.000003 seconds to execute 
Element 200000 found and returned.
fun() took 0.000003 seconds to execute 
Element 400000 found and returned.
fun() took 0.000004 seconds to execute 
Element 800000 found and returned.
fun() took 0.000003 seconds to execute

Answer 2

您的计时方法非常规。通常，要检查算法性能，

取几次尝试的平均值。
计时几个不同大小的容器。

在伪代码中：

For N in [2..9]:
  Fill container with 10^N items
  lookupTime = 0;
  generate M values in range
  for i in [1..M]:
    lookupTime += duration(lookup(value[i]))
  performance[N] = lookupTime/M;

我的跳过列表是否真的在N而不是log（N）中搜索？

2 个答案: