C++primer Chapter 9 Sequence Container After-Class Answer (Continuous Update)

9.1

(a) list
(b) deque, if you don’t entangle the constant, then list can be
(c) vector

9.2

list<deque<int>> ldi;

9.3

For satisfying the following conditions, two iterators begin and end form an iterator range:

• They point to an element in the same container, or a position after the last element of the container
• We can reach end by repeatedly incrementing begin, that is, end is not before begin

9.4

#include<iostream>
#include<vector>

using namespace std;

typedef vector<int>::iterator veci;
bool find(veci A, veci B, int c) {
    
    
	while(A != B) {
    
    
		if(*A == c) return true;
		++A;
	}
	return false;
}

int main()
{
    
    
	vector<int> vec = {
    
    9, 9, 8, 2, 4, 4, 3, 5, 3};
	for(int i = 0; i < 10; ++i) {
    
    
		if(find(vec.begin(), vec.end(), i)) cout << i << " is in vec.\n";
		else cout << i << " is not in vec\n";
	}
	return 0;
}

9.5

#include<iostream>
#include<vector>

using namespace std;

typedef vector<int>::iterator veci;
veci find(veci A, veci B, int c) {
    
    
	while(A != B) {
    
    
		if(*A == c) return A;
		++A;
	}
	return B;
}

int main()
{
    
    
	vector<int> vec = {
    
    9, 9, 8, 2, 4, 4, 3, 5, 3};
	for(int i = 0; i < 10; ++i) {
    
    
		auto it = find(vec.begin(), vec.end(), i);
		if(it == vec.end()) cout << i << " is not in vec.\n";
		else cout << *it << " is in vec\n";
	}
	return 0;
}

9.6

list<int> lst1;
list<int>::iterator iter1 = lst1.begin(), 
					iter2 = lst1.end();
while(iter1 < iter2) /* ... */

The arithmetic operators of iterators can only be applied to iterators of string, vector, deque, and array.
Therefore need to

while(iter1 < iter2) 

change into:

while(iter1 != iter2)