[Data structure]--bucket sequencing

One, bucket sort

The following code goes from: Bucket sorting

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2. #include <iostream>
4. #include <list>
8. using namespace std;
12. struct Node
14. {
16. Double value;
18. Node *next;
20. };
22. //Bucket sequencing main program
24. void Bucketsort (double* arr, int length)
26. {
28. Node KEY[10];
30. int number = 0;
32. Node *p, *q; //Insert Node TEMP variable
34. int counter = 0;
36. For (int i = 0; i <; i++)
38. {
40. Key[i].value = 0;
42. Key[i].next = NULL;
44. }
48. For (int i = 0; i < length; i++)
50. {
52. Node *insert = new node ();
54. Insert->value = Arr[i];
56. Insert->next = NULL;
58. Number = arr[i] * 10;
60. if (Key[number].next = = NULL)
62. {
64. Key[number].next = insert;
66. }
68. Else
70. {
72. p = &key[number];
74. Q = key[number].next;
76. While ((q! = NULL) && (Q->value <= arr[i]))
78. {
80. Q = q->next;
82. p = p->next;
84. }
86. Insert->next = q;
88. P->next = insert;
90. }
92. }
94. For (int i = 0; i <; i++)
96. {
98. p = key[i].next;
100. if (p = = NULL)
102. continue;
104. While (P! = NULL)
106. {
108. arr[counter++] = p->value;
110. p = p->next;
112. }
114. }
116. }
120. int main ()
122. {
124. Double a[] = {0.78, 0.17, 0.39, 0.26, 0.72, 0.94, 0.21, 0.12, 0.23, 0.68};
126. Bucketsort (A, 10);
128. For (int i = 0; i <; i++)
130. {
132. cout << A[i] << "";
134. }
136. cout << Endl;
138. return 0;
140. }

"Example" to a mix of 52 playing cards by the number of a<2<...<j<q<k sort, you need to set 13 "box", sorting in turn each card by the number of points into the corresponding box, and then in turn these boxes end-to-end, and then get the number of points ascending order of a pair
In general, how many keywords in each box are stored in the same record is unpredictable, so the type of box should be designed as a linked list is appropriate.

[Data structure]--bucket sequencing