C++ primer 9.3.1节练习

9.18 - 9.22

//***********9.3.1节练习 9.18**************
string temp;
deque<string> sdeq;//默认初始化
while (cin >> temp)
{
    sdeq.push_back(temp);
}
for (auto i = sdeq.cbegin(); i != sdeq.cend(); ++i)
    cout << *i << endl;


//***********9.19************************
list<string> slist;
string temp1;
while (cin >> temp1)
{
    slist.push_back(temp1);
}
for (auto i = slist.cbegin(); i != slist.cend(); ++i)
    cout << *i << endl;
//故和上一个练习相比，并无太多改动


//*************9.20***********************
list<int> ilist;
for (int i = 0; i != 10; ++i)
{
    ilist.push_back(i);
}
deque<int> odddeq,evendeq;
for (auto i = ilist.cbegin(); i != ilist.cend(); ++i)
{
    if (*i & 1 )//用逻辑位与运算，和%2等价
        odddeq.push_back(*i);
    else
        evendeq.push_back(*i);
}

//**************9.21**********************

vector<string> svec;
string word;
auto iter = svec.begin();
while (cin >> word)
{
    iter = svec.insert(iter, word);
}
for (auto i : svec)
    cout << i << endl;
//此操作相当于一直push_front，从而元素是倒序的

std::system("pause");
return 0;  

//*****************9.22练习***********************

//目的是检查，前一半元素中如果有和some_val相等的元素的，就在他前面插入一个2*some_val
//而错误程序是，当插入元素时，原本的迭代器会失效，所以我们要更新这个mid，让他始终在最初始的位置

vector<int> test = { 1, 2, 1, 3, 1, 5, 2, 1, 4, 1 };
for (auto i : test)
{
    cout << i << " ";
}

cout << endl;

int some_val = 1;
int newsum = 0;//用来记录插入的新元素的个数
size_t org_size = test.size();
vector<int>::iterator iter = test.begin();

while (iter != test.begin() + org_size / 2 + newsum)// !=后面是新的mid位置
{
    if (*iter == some_val)
    {
        iter = test.insert(iter, 2 * some_val);
        iter += 2;
        newsum++;
    }
    else
        iter++;
}

for (auto i : test)
{
    cout << i << " ";
}

cout << endl;

std::system("pause");
return 0; 

 

来源：https://www.cnblogs.com/Kanna/p/12268433.html