C++算法 merge()函数用于将两个排序参数[first1,last1)和[first2,last2)合并为一个从结果开始的排序参数。
default(1) template <class InputIterator1, class InputIterator2, class OutputIterator> OutputIterator merge (InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result); custom (2) template <class InputIterator1, class InputIterator2, class OutputIterator, class Compare> OutputIterator merge (InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result, Compare comp);
first1 :一个输入迭代器,它指向要合并的第一个排序的源序列中的第一个元素。
last:一个输入迭代器,它指向要合并的第一个排序源序列中的最后一个元素。
first2 :一个输入迭代器,它指向要合并的第二个已排序源序列中的第一个元素。
last2 :一个输入迭代器,它指向要合并的第二个已排序源序列中的最后一个元素。
comp :用户定义的二进制谓词函数,该函数接受两个参数,如果两个参数顺序正确,则返回true,否则返回false。
val :比较参数内元素的上限值。
result:一个输出迭代器,该迭代器指向目标参数中的第一个元素,其中两个源参数将合并为一个排序参数。
它返回一个指向结果序列中最后一个元素的迭代器。
让我们看一个简单的示例来演示merge()的用法:
#include <iostream> #include <algorithm> #include <vector> using namespace std; void printVector(vector<int>& v) { for (vector<int>::iterator it = v.begin(); it != v.end(); ++it) cout << ' ' << *it; cout << '\n'; } int main () { vector<int> v1 = {5,1,4,2,6}, v2 = {50,40,30,20,10}, v3(10); sort(v1.begin(), v1.end()); sort(v2.begin(), v2.end()); merge(v1.begin(),v1.end(),v2.begin(),v2.end(),v3.begin()); cout << "Vector v1 : "; printVector(v1); cout << "Vector v2 : "; printVector(v2); cout << "Vector v3 : "; printVector(v3); return 0; }
输出:
Vector v1 : 1 2 4 5 6 Vector v2 : 10 20 30 40 50 Vector v3 : 1 2 4 5 6 10 20 30 40 50
我们来看另一个简单的示例e,该示例使用operator <来实现merge()函数
#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { //initializing 1st container vector<int> arr1 = { 1, 4, 6, 3, 2 }; //initializing 2nd container vector<int> arr2 = { 60, 20, 50, 70, 10 }; //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; }
输出:
The container after merging initial containers is: 1 2 3 4 6 10 20 50 60 70
让我们看另一个简单的示例,使用比较函数演示merge()
#include <iostream> #include <vector> #include <algorithm> 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 = { 60, 20, 50, 70, 10 }; //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; }
输出:
The container after reverse merging initial containers is : 70 60 50 20 10 6 4 3 2 1
我们来看另一个简单的例子
#include <iostream> #include <algorithm> #include <vector> using namespace std; int main() { //initializing 1st container //containing denominations vector<int> stack1 = { 50, 20, 10, 100, 200 }; //initializing 2nd container //containing demonitions vector<int> stack2 = { 500, 2000, 5000, 1000, 10000 }; //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; }
输出:
The original 1st stack: 50 20 10 100 200 The original 2nd stack: 500 2000 5000 1000 10000 The resultant stack of notes is: 10 20 50 100 200 500 1000 2000 5000 10000
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