C++中的STL库提供的合并()，这是非常有用的排序两个容器合并成一个单一的容器中。
它在标题“ algorithm ”中定义。它以两种方式实现。
语法1：使用运算符“ <”
Template :
template
outiter merge (initer1 beg1, initer1 end1,
initer2 beg2, initer2 end2,
outiter res)
Parameters :
beg1 : Input iterator to initial position of first sequence.
end1 : Input iterator to final position of first sequence.
beg2 : Input iterator to initial position of second sequence.
end2 : Input iterator to final position of second sequence.
res : Output Iterator to initial position of resultant container.
Return value :
Iterator to last element of the resulting container.
// C++ code to demonstrate the working of
// merge() implementation 1
#include
using namespace std;
int main()
{
// initializing 1st container
vector arr1 = { 1, 4, 6, 3, 2 };
// initializing 2nd container
vector arr2 = { 6, 2, 5, 7, 1 };
// declaring resultant container
vector arr3(10);
// sorting initial containers
sort(arr1.begin(), arr1.end());
sort(arr2.begin(), arr2.end());
// using merge() to merge the initial containers
merge(arr1.begin(), arr1.end(), arr2.begin(), arr2.end(), arr3.begin());
// printing the resultant merged container
cout << "The container after merging initial containers is : ";
for (int i = 0; i < arr3.size(); i++)
cout << arr3[i] << " ";
return 0;
}
输出：
The container after merging initial containers is : 1 1 2 2 3 4 5 6 6 7
语法2：使用比较器函数
Template :
template
outiter merge (initer1 beg1, initer1 end1,
initer2 beg2, initer2 end2,
outiter res, Compare comp)
Parameters :
beg1 : Input iterator to initial position of first sequence.
end1 : Input iterator to final position of first sequence.
beg2 : Input iterator to initial position of second sequence.
end2 : Input iterator to final position of second sequence.
res : Output Iterator to initial position of resultant container.
comp : The comparator function that returns a boolean
true/false of the each elements compared. This function
accepts two arguments. This can be function pointer or
function object and cannot change values.
Return value :
Iterator to last element of the resulting container.
// C++ code to demonstrate the working of
// merge() implementation 2
#include
using namespace std;
// comparator function to reverse merge sort
struct greaters {
bool operator()(const long& a, const long& b) const
{
return a > b;
}
};
int main()
{
// initializing 1st container
vector arr1 = { 1, 4, 6, 3, 2 };
// initializing 2nd container
vector arr2 = { 6, 2, 5, 7, 1 };
// declaring resultant container
vector arr3(10);
// sorting initial containers
// in descending order
sort(arr1.rbegin(), arr1.rend());
sort(arr2.rbegin(), arr2.rend());
// using merge() to merge the initial containers
// returns descended merged container
merge(arr1.begin(), arr1.end(), arr2.begin(), arr2.end(), arr3.begin(), greaters());
// printing the resultant merged container
cout << "The container after reverse merging initial containers is : ";
for (int i = 0; i < arr3.size(); i++)
cout << arr3[i] << " ";
return 0;
}
输出 ：
The container after reverse merging initial containers is : 7 6 6 5 4 3 2 2 1 1
可能的应用：合并函数可用于按排序顺序提供两个堆栈的单个堆栈。这些可以是一堆书或笔记。让我们讨论一个简单的示例，该示例根据其值将升序的两叠钞票合并为一个。
// C++ code to demonstrate the application of
// merge() stacking notes
#include
using namespace std;
int main()
{
// initializing 1st container
// containing denominations
vector stack1 = { 50, 20, 10, 100, 2000 };
// initializing 2nd container
// containing demonitions
vector stack2 = { 500, 2000, 10, 100, 50 };
// declaring resultant stack
vector stack3(10);
cout << "The original 1st stack : ";
for (int i = 0; i < 5; i++)
cout << stack1[i] << " ";
cout << endl;
cout << "The original 2nd stack : ";
for (int i = 0; i < 5; i++)
cout << stack2[i] << " ";
cout << endl;
// sorting initial stacks of notes
// in descending order
sort(stack1.begin(), stack1.end());
sort(stack2.begin(), stack2.end());
// using merge() to merge the initial stacks
// of notes
merge(stack1.begin(), stack1.end(), stack2.begin(), stack2.end(), stack3.begin());
// printing the resultant stack
cout << "The resultant stack of notes is : ";
for (int i = 0; i < stack3.size(); i++)
cout << stack3[i] << " ";
return 0;
}
输出 ：
The original 1st stack : 50 20 10 100 2000
The original 2nd stack : 500 2000 10 100 50
The resultant stack of notes is : 10 10 20 50 50 100 100 500 2000 2000
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