Simple hash 1

Description

Using a linear probe method ( Linear probin is unused G ) can resolve the conflict in the hash, the basic idea is: a hash function set H (Key) = D , and assuming the hash memory structure is a circular array , then when conflict occurs when , continue probing d + 1, d + 2 , ... , until the conflict is resolved .

For example , an existing key set is {47 , 7 , 29 , 11 , 16 , 92 , 22 , 8 , 3} , is provided for the hash table entry length m = 11 , the hash function Hash (key) = key mod 11 , linear conflict detection method.

Build a hash table is as follows:

0	1	2	3	4	5	6	7	8	9	10
11	22		47	92	16	3	7	29	8

Now given hash function to the Hash (Key) Key = MOD m , according to the above rules required , the use of the linear probe method conflict . Required to establish the corresponding hash table, as required for printing.

Input

Only one test case , the first one acts integers n and m ( 1 <= n, m <= 10 , 000 ), n the representative key of the total number , m represents the length of the hash table , and the hash function is to make the Hash (key ) Key = MOD m .

Next n lines, each line an integer representing a key , wherein K EY and key pairwise identical ( 0 <= key <= 10,000 ) .

Output

Output establish good hash table, such as the following table :

0	1	2	3	4	5	6	7	8	9	10
11	22		47	92	16	3	7	29	8

Output should

0#11

1#22

2#NULL

3#47

4#92

5#16

6#3

7#7

8#29

9#8

10#NULL

Sample Input

Sample Output

0#5
1#1
2#6
3#NULL
4 # NULL 

directly on the code:

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int hash(int key, int len) {
 5     return (key % len);
 6 }
 7 
 8 int hashTable[10000];
 9 
10 void initializeHash(int size, int lenOfHash) {
11     int temp;
12     
13     for (int i = 0; i < size; i++) {
14         cin >> temp;
15         
16         if (hashTable[hash(temp, lenOfHash)] == -1) {
17             hashTable[hash(temp, lenOfHash)] = temp;
18         } else {
19             int pos = hash(temp, lenOfHash);
20             while (hashTable[pos] != -1) {
21                 pos = (pos + 1) % lenOfHash;
22             }
23             
24             hashTable[pos] = temp;
25         }
26     }
27 }
28 
29 void print(int lenOfHash) {
30     for (int i = 0; i < lenOfHash; i++) {
31         if (hashTable[i] != -1) {
32             cout << i << "#" << hashTable[i] << endl;
33         } else {
34             cout << i << "#NULL" << endl;
35         }
36     }
37 }
38 
39 int main()
40 {
41     int numOfKey, lenOfHash;
42     
43     cin >> numOfKey;
44     cin >> lenOfHash;
45 
46     for (int i = 0; i < lenOfHash; i++) {
47         hashTable[i] = -1;
48     }
49     
50     initializeHash(numOfKey, lenOfHash);
51     
52     print(lenOfHash);
53     
54     return 0;
55 }

Reproduced in: https: //www.cnblogs.com/IT-nerd/p/3462370.html

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