给定数组arr的大小N ，任务是当数组的最大和第二个最大元素被数组中的绝对差值重复替换时，打印数组中剩余的最终数组值。
注意：如果最多两个元素相同，则将从数组中删除这两个元素，而不替换任何值。
例子：
Input: arr = [2, 7, 4, 1, 8, 1]
Output: 1
Explanations:
Merging 7 and 8: absolute difference = 7 – 8 = 1. So the array converted into [2, 4, 1, 1, 1].
Merging 2 and 4: absolute difference = 4 – 2 = 2. So the array converted into [2, 1, 1, 1].
Merging 2 and 1: absolute difference = 2 – 1 = 1. So the array converted into [1, 1, 1].
Merging 1 and 1: absolute difference = 4 – 2 = 0. So nothing will be Merged.
So final array = [1].
Input: arr = [7, 10, 5, 4, 11, 25]
Output: 2
高效方法：使用优先级队列
- 设置一个优先级队列(二进制最大堆)，该队列自动按排序顺序排列元素。
- 然后选择第一个元素(最大)和第二个元素(第二个最大)，如果两者相等则不推送任何内容，如果不相等则按队列中两者的绝对差进行推送。
- 执行上述步骤，直到队列大小等于1，然后返回最后一个元素。如果队列在达到大小1之前变空，则返回0。
下面是上述方法的实现：
C++
// C++ program to find the array value
// by repeatedly replacing max 2 elements
// with their absolute difference
#include
using namespace std;
// function that return last
// value of array
int lastElement(vector& arr)
{
// Build a binary max_heap.
priority_queue pq;
for (int i = 0; i < arr.size(); i++) {
pq.push(arr[i]);
}
// For max 2 elements
int m1, m2;
// Iterate until queue is not empty
while (!pq.empty()) {
// if only 1 element is left
if (pq.size() == 1)
// return the last
// remaining value
return pq.top();
m1 = pq.top();
pq.pop();
m2 = pq.top();
pq.pop();
// check that difference
// is non zero
if (m1 != m2)
pq.push(m1 - m2);
}
// finally return 0
return 0;
}
// Driver Code
int main()
{
vector arr = { 2, 7, 4, 1, 8, 1, 1 };
cout << lastElement(arr) << endl;
return 0;
} Java
// Java program to find the array value
// by repeatedly replacing max 2 elements
// with their absolute difference
import java.util.*;
class GFG{
// Function that return last
// value of array
static int lastElement(int[] arr)
{
// Build a binary max_heap
PriorityQueue pq = new PriorityQueue<>(
(a, b) -> b - a);
for(int i = 0; i < arr.length; i++)
pq.add(arr[i]);
// For max 2 elements
int m1, m2;
// Iterate until queue is not empty
while(!pq.isEmpty())
{
// If only 1 element is left
if (pq.size() == 1)
{
// Return the last
// remaining value
return pq.poll();
}
m1 = pq.poll();
m2 = pq.poll();
// Check that difference
// is non zero
if (m1 != m2)
pq.add(m1 - m2);
}
// Finally return 0
return 0;
}
// Driver code
public static void main(String[] args)
{
int[] arr = new int[]{2, 7, 4, 1, 8, 1, 1 };
System.out.println(lastElement(arr));
}
}
// This code is contributed by dadi madhav Python3
# Python3 program to find the array value
# by repeatedly replacing max 2 elements
# with their absolute difference
from queue import PriorityQueue
# Function that return last
# value of array
def lastElement(arr):
# Build a binary max_heap.
pq = PriorityQueue()
for i in range(len(arr)):
# Multipying by -1 for
# max heap
pq.put(-1 * arr[i])
# For max 2 elements
m1 = 0
m2 = 0
# Iterate until queue is not empty
while not pq.empty():
# If only 1 element is left
if pq.qsize() == 1:
# Return the last
# remaining value
return -1 * pq.get()
else:
m1 = -1 * pq.get()
m2 = -1 * pq.get()
# Check that difference
# is non zero
if m1 != m2 :
pq.put(-1 * abs(m1 - m2))
return 0
# Driver Code
arr = [ 2, 7, 4, 1, 8, 1, 1 ]
print(lastElement(arr))
# This code is contributed by ishayadav181C#
// C# program to find the array value
// by repeatedly replacing max 2 elements
// with their absolute difference
using System;
using System.Collections.Generic;
class GFG{
// Function that return last
// value of array
static int lastElement(int[] arr)
{
// Build a binary max_heap
Queue pq = new Queue();
for(int i = 0; i < arr.Length; i++)
pq.Enqueue(arr[i]);
// For max 2 elements
int m1, m2;
// Iterate until queue is not empty
while (pq.Contains(0))
{
// If only 1 element is left
if (pq.Count == 1)
{
// Return the last
// remaining value
return pq.Peek();
}
m1 = pq.Dequeue();
m2 = pq.Peek();
// Check that difference
// is non zero
if (m1 != m2)
pq.Enqueue(m1 - m2);
}
// Finally return 0
return 0;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 2, 7, 4, 1, 8, 1, 1 };
Console.WriteLine(lastElement(arr));
}
}
// This code is contributed by sanjoy_62 输出：
0时间复杂度： O(N)
辅助复杂度： O(N)