数据结构算法C语言实现（十二）--- 3.4循环队列&队列的顺序表示和实现

 1 //3_4_part1.h
 2 /**
 3 author:zhaoyu
 4 email:zhaoyu1995.com@gmail.com
 5 date:2016-6-9
 6 note:realize my textbook <<数据结构（C语言版）>>
 7 */
 8 //Page 64
 9 #include <cstdio>
10 #include "head.h"
11 #define QElemType int
12 //----循环队列：队列的顺序存储结构----
13 #define MAXQSIZE 10 //最大队列长度
14 typedef struct{
15     QElemType *base;
16     int front;
17     int rear;
18 }SqQueue;
19 //----循环队列的基本操作说明及实现----
20 Status InitQueue(SqQueue &Q)
21 {
22     //构造一个空队列 Q
23     Q.base = (QElemType *)malloc(MAXQSIZE*sizeof(QElemType));
24     if (!Q.base)
25     {
26         exit(OVERFLOW);
27     }
28     Q.front = Q.rear = 0;
29     return OK;
30 }
31 int QueueLength(SqQueue Q)
32 {
33     //返回 Q 的元素个数，即队列的长度
34     return (Q.rear - Q.front + MAXQSIZE) % MAXQSIZE;
35 }
36 Status EnQueue(SqQueue &Q, QElemType e)
37 {
38     //插入元素 e 为 Q 的新的队尾元素
39     if ((Q.rear+1)%MAXQSIZE == Q.front)
40     {
41         return ERROR;//队列满
42     }
43     Q.base[Q.rear] = e;
44     Q.rear = (Q.rear+1)%MAXQSIZE;
45     return OK;
46 }
47 Status DeQueue(SqQueue &Q, QElemType &e)
48 {
49     //若队列不空，则删除 Q 的队列头元素，用 e 返回其值，
50     //并返回 OK，否则返回 ERROR
51     if (Q.front == Q.rear)
52     {
53         return ERROR;
54     }
55     e = Q.base[Q.front];
56     Q.front = (Q.front+1)%MAXQSIZE;
57     return OK;
58 }
59 void PrintQueue(SqQueue Q)
60 {
61     int cnt = Q.front;
62     if (Q.front == Q.rear)
63     {
64         printf("void\n");
65         return;
66     }
67     while ((cnt+1)%MAXQSIZE != Q.rear)
68     {    
69         //printf("%d\t%d\n",Q.base[cnt++], cnt);输出好奇怪
70         printf("%d\t", Q.base[cnt]);
71         cnt++;
72     }
73     printf("%d\n", Q.base[cnt]);
74 }

 1 #include "3_4_part2.h"
 2 int main(int argc, char const *argv[])
 3 {
 4     SqQueue Q;
 5     InitQueue(Q);
 6     for (int i = 1; i < 10; ++i)
 7     {
 8         EnQueue(Q, i*10);
 9     }
10     PrintQueue(Q);
11     int e;
12     EnQueue(Q, 100);
13     PrintQueue(Q);
14     DeQueue(Q, e);
15     printf("%d\n", e);
16     EnQueue(Q, 100);
17     PrintQueue(Q);
18     return 0;
19 }

 1 #include <stdio.h>
 2 int main(int argc, char const *argv[])
 3 {
 4     int cnt = 0, cnt_1 = 0, cnt_2 = 0, cnt_3 = 0, cnt_4 = 0;
 5     int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
 6     for (int i = 1; i <= 4; i++)
 7     {
 8         printf("%d-", cnt);
 9         printf("%d\t", a[cnt++]);
10     }
11     printf("\n");
12     for (int i = 1; i <= 4; i++)
13     {
14         printf("%d-%d\t", a[cnt_1++], cnt_1);
15     }
16     printf("\n");
17     for (int i = 1; i <= 4; i++)
18     {
19         printf("%d-%d\t", a[++cnt_2], cnt_2);
20     }
21     printf("\n");
22     for (int i = 1; i <= 4; i++)
23     {
24         printf("%d-%d\t", cnt_3, a[cnt_3++]);
25     }
26     printf("\n");
27     for (int i = 1; i <= 4; i++)
28     {
29         printf("%d-%d\t", cnt_4, a[++cnt_4]);
30     }
31     printf("\n");
32     int cnt_5 = 1, cnt_6 = 1, cnt_7 = 1;
33     printf("%d-%d\n", cnt_5, 5*(cnt_5++));
34     return 0;
35 }