C++在其STL庫中提供了merge()，它對於將排序兩個容器合並為一個容器非常有用。它在標題“算法”中定義。它以兩種方式實現。
語法1：使用運算符“ <”
Template:
template
outiter merge (initer1 beg1, initer1 end1,
initer2 beg2, initer2 end2,
outiter res)
參數：
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.
返回值:
Iterator to last element of the resulting container.
// C++ code to demonstrate the working of
// merge() implementation 1
#include <bits/stdc++.h>
using namespace std;
int main()
{
// initializing 1st container
vector<int> arr1 = { 1, 4, 6, 3, 2 };
// initializing 2nd container
vector<int> arr2 = { 6, 2, 5, 7, 1 };
// declaring resultant container
vector<int> 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;
}輸出：
The container after merging initial containers is:1 1 2 2 3 4 5 6 6 7
語法2：使用比較器函數
Template:
template
outiter merge (initer1 beg1, initer1 end1,
initer2 beg2, initer2 end2,
outiter res, Compare comp)
參數：
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.
返回值:
Iterator to last element of the resulting container.
// C++ code to demonstrate the working of
// merge() implementation 2
#include <bits/stdc++.h>
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<int> arr1 = { 1, 4, 6, 3, 2 };
// initializing 2nd container
vector<int> arr2 = { 6, 2, 5, 7, 1 };
// declaring resultant container
vector<int> 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;
}輸出：
The container after reverse merging initial containers is:7 6 6 5 4 3 2 2 1 1
可能的應用：合並函數可用於按排序順序提供兩個堆棧的單個堆棧。這些可以是一堆書或筆記。讓我們討論一個簡單的示例，該示例根據其值將升序的兩疊鈔票合並為一個。
// C++ code to demonstrate the application of
// merge() stacking notes
#include <bits/stdc++.h>
using namespace std;
int main()
{
// initializing 1st container
// containing denominations
vector<int> stack1 = { 50, 20, 10, 100, 2000 };
// initializing 2nd container
// containing demonitions
vector<int> stack2 = { 500, 2000, 10, 100, 50 };
// declaring resultant stack
vector<int> 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;
}輸出：
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
相關用法
注：本文由純淨天空篩選整理自 merge() in C++ STL。非經特殊聲明，原始代碼版權歸原作者所有，本譯文未經允許或授權，請勿轉載或複製。