给定一个数组arr大小N ，任务是打印数组中剩余的最终数组值，当数组的最大和第二个最大元素被替换为它们在数组中的绝对差时，重复。
注意：如果最大两个元素相同，则将两者从数组中删除，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：使用优先队列
- 制作一个优先队列(二进制最大堆)，它会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，如果不相等则推送队列中两者的绝对差异。
- 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)
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