Big Java, Binder Ready Version: Early Objects
6th Edition
ISBN: 9781119056447
Author: Cay S. Horstmann
Publisher: WILEY
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Chapter 19.3, Problem 13SC
Program Plan Intro
Stream collectors:
- Collectors are an alternative approach to reductions.
- It is used to gather stream elements into collection which performs other types of reductions.
- The interface “Stream <T>” has a method which uses an object in order to reduce a stream of elements into a single value.
- The package “Stream” contains a class named “Collectors” which offers many static methods that return various types of collectors.
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Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head =
{
NULL) return 0;
first;
while (head->next != NULL) {
}
if (head
first) {
return 1; }
head = head->next;
return 0;
void test ll_has_cycle () {
int i;
node nodes [6];
for (i = 0; i < 6; i++) {
nodes [i] .next = NULL;
nodes [i].value = i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next
=
&nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next
nodes [4] .next
&nodes [4];
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
how to read log logs
Discrete Mathematics for Computer Engineering
Chapter 19 Solutions
Big Java, Binder Ready Version: Early Objects
Ch. 19.1 - Prob. 1SCCh. 19.1 - Prob. 2SCCh. 19.1 - Prob. 3SCCh. 19.1 - Prob. 4SCCh. 19.1 - Prob. 5SCCh. 19.1 - Prob. 6SCCh. 19.2 - Prob. 7SCCh. 19.2 - Prob. 8SCCh. 19.2 - Prob. 9SCCh. 19.2 - Prob. 10SC
Ch. 19.3 - Prob. 11SCCh. 19.3 - Prob. 12SCCh. 19.3 - Prob. 13SCCh. 19.3 - Prob. 14SCCh. 19.3 - Prob. 15SCCh. 19.4 - Prob. 16SCCh. 19.4 - Prob. 17SCCh. 19.4 - Prob. 18SCCh. 19.4 - Prob. 19SCCh. 19.4 - Prob. 20SCCh. 19.5 - Prob. 21SCCh. 19.5 - Prob. 22SCCh. 19.5 - Prob. 23SCCh. 19.5 - Prob. 24SCCh. 19.5 - Prob. 25SCCh. 19.6 - Prob. 26SCCh. 19.6 - Prob. 27SCCh. 19.6 - Prob. 28SCCh. 19.6 - Prob. 29SCCh. 19.6 - Prob. 30SCCh. 19.7 - Prob. 31SCCh. 19.7 - Prob. 32SCCh. 19.7 - Prob. 33SCCh. 19.7 - Prob. 34SCCh. 19.7 - Prob. 35SCCh. 19.8 - Prob. 36SCCh. 19.8 - Prob. 37SCCh. 19.8 - Prob. 38SCCh. 19.8 - Prob. 39SCCh. 19.9 - Prob. 40SCCh. 19.9 - Prob. 41SCCh. 19.9 - Prob. 42SCCh. 19.9 - Prob. 43SCCh. 19.9 - Prob. 44SCCh. 19.10 - Prob. 45SCCh. 19.10 - Prob. 46SCCh. 19.10 - Prob. 47SCCh. 19.10 - Prob. 48SCCh. 19.10 - Prob. 49SCCh. 19 - Prob. 1RECh. 19 - Prob. 2RECh. 19 - Prob. 3RECh. 19 - Prob. 4RECh. 19 - Prob. 5RECh. 19 - Prob. 6RECh. 19 - Prob. 7RECh. 19 - Prob. 8RECh. 19 - Prob. 9RECh. 19 - Prob. 10RECh. 19 - Prob. 11RECh. 19 - Prob. 12RECh. 19 - Prob. 13RECh. 19 - Prob. 1PECh. 19 - Prob. 2PECh. 19 - Prob. 3PECh. 19 - Prob. 4PECh. 19 - Prob. 5PECh. 19 - Prob. 6PECh. 19 - Prob. 7PECh. 19 - Prob. 8PECh. 19 - Prob. 9PECh. 19 - Prob. 10PECh. 19 - Prob. 11PECh. 19 - Prob. 12PECh. 19 - Prob. 13PECh. 19 - Prob. 14PECh. 19 - Prob. 15PECh. 19 - Prob. 16PECh. 19 - Prob. 17PECh. 19 - Prob. 18PECh. 19 - Prob. 19PECh. 19 - Prob. 20PECh. 19 - Prob. 1PPCh. 19 - Prob. 2PPCh. 19 - Prob. 3PPCh. 19 - Prob. 4PPCh. 19 - Prob. 5PPCh. 19 - Prob. 6PPCh. 19 - Prob. 7PPCh. 19 - Prob. 8PPCh. 19 - Prob. 9PPCh. 19 - Prob. 10PPCh. 19 - Prob. 11PPCh. 19 - Prob. 12PP
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