使用C ++中的另一个数组最大化元素
问题陈述
给定两个大小为n的数组，请使用第二个数组中的元素来最大化第一个数组，以使形成的新数组包含两个数组中n个最大但唯一的元素，赋予第二个数组优先级，即第二个数组的所有元素都出现在第一个数组。元素的出现顺序在输出中应与输入中的相同
如果arr1[]={12，15，10}和arr2[]={16，17，5}，则{16，17，15}是通过维持顺序来自两个数组的最大元素。
算法
1. Create temporary array of size 2 * n 2. Store elements of arr1 and arr2 in temporary array and sort it in descending order 3. To keep the order of elements according to input arrays we will use hash table 4. Store first n largest unique elements of temporary array in hash table 5. Traverse the second array and store that elements of second array in temporary array that are present in hash table 6. Similarly, traverse the first array and store the elements that are present in hash table 7. In this way we get n unique and largest elements from both the arrays in temporary array
示例
#include <bits/stdc++.h>
using namespace std;
void printArray(int *arr, int n){
for (int i = 0; i < n; ++i) {
cout << arr[i] << " ";
}
cout << endl;
}
bool compare(int a, int b){
return a > b;
}
void getMaxElements(int *arr1, int *arr2, int n){
int temp[2 * n];
int k = 0;
for (int i = 0; i < n; ++i) {
temp[k] = arr1[i];
++k;
}
for (int i = 0; i < n; ++i) {
temp[k] = arr2[i];
++k;
}
sort(temp, temp + 2 * n, compare);
unordered_set<int> hash;
int i = 0;
while (hash.size() != n) {
if (hash.find(temp[i]) == hash.end()) {
hash.insert(temp[i]);
}
++i;
}
k = 0;
for (int i = 0; i < n; ++i) {
if (hash.find(arr2[i]) != hash.end()) {
temp[k++] = arr2[i];
hash.erase(arr2[i]);
}
}
for (int i = 0; i < n; ++i) {
if (hash.find(arr1[i]) != hash.end()) {
temp[k++] == arr1[i];
hash.erase(arr1[i]);
}
}
for (int i = 0; i < n; ++i) {
arr1[i] = temp[i];
}
}
int main(){
int arr1[] = {12, 15, 10};
int arr2[] = {16, 17, 5};
int n = sizeof(arr1) / sizeof(arr1[0]);
cout << "First array:\n";
printArray(arr1, n);
cout << "Second array:\n";
printArray(arr2, n);
getMaxElements(arr1, arr2, n);
cout << "Maximum array:\n";
printArray(arr1, n);
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
First array: 12 15 10 Second array: 16 17 5 Maximum array: 16 17 15