请有人向我解释这个问题的时间复杂度使用最小堆

问题描述

这是代码的链接。我们如何获得 O(N*logM + M*logN) 的时间复杂度。我是初学者请尽量详细解释。

来源：https://www.geeksforgeeks.org/merge-two-sorted-arrays-in-constant-space-using-min-heap/#:~:text=Start%20traversing%20the%20first%20array,the%20root%20of%20the%20min_heap。

// C++ program to merge two sorted arrays in
// constant space
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to merge two sorted arrays in
// constant space
void mergeArrays(int* a,int n,int* b,int m)
{
  
    // Convert second array into a min_heap
    // using make_heap() STL function [takes O(m)]
    make_heap(b,b + m,greater<int>());
  
    // Start traversing the first array
    for (int i = 0; i < n; i++) {
  
        // If current element is greater than root
        // of min_heap
        if (a[i] > b[0]) {
  
            // Pop minimum element from min_heap using
            // pop_heap() STL function
            // The pop_heap() function removes the minimum element from
            // heap and moves it to the end of the container
            // converted to heap and reduces heap size by 1
            pop_heap(b,greater<int>());
  
            // Swapping the elements
            int tmp = a[i];
            a[i] = b[m - 1];
            b[m - 1] = tmp;
  
            // Apply push_heap() function on the container
            // or array to again reorder it in the
            // form of min_heap
            push_heap(b,greater<int>());
        }
    }
  
    // Convert the second array again into max_heap
    // because sort_heap() on min heap sorts the array
    // in decreasing order
    // This step is [O(m)]
    make_heap(b,b + m); // It's a max_heap
  
    // Sort the second array using sort_heap() function
    sort_heap(b,b + m);
}
  
// Driver Code
int main()
{
  
    int ar1[] = { 1,5,9,10,15,20 };
    int ar2[] = { 2,3,8,13 };
    int m = sizeof(ar1) / sizeof(ar1[0]);
    int n = sizeof(ar2) / sizeof(ar2[0]);
    mergeArrays(ar1,m,ar2,n);
  
    cout << "After Merging :- \nFirst Array: ";
    for (int i = 0; i < m; i++)
        cout << ar1[i] << " ";
    cout << "\nSecond Array: ";
    for (int i = 0; i < n; i++)
        cout << ar2[i] << " ";
  
    return 0;
}

解决方法

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