给定大小为N的数组arr [] ，任务是在最长的子数组中找到仅由偶数或奇数组成的最大元素。
例子：
Input: arr[] = { 2, 4, 6, 9, 10, 11 }
Output: 6
Explanation:
The longest subarray consisting of only even or odd numbers is { arr[0], arr[1], arr[2] }.
Since the largest element of the subarray is arr[2], the required output is 6.
Input: arr[] = { 3, 5, 7, 4, 9, 11, 13 }
Output: 13
Explanation:
The longest subarray consisting of only even or odd numbers are { {3, 5, 7 }, { 9, 11, 13 } }.
The largest elements in the subarrays are 7 and 13 respectively. 13 being the largest, is the required output.
方法：请按照以下步骤解决问题：
- 初始化一个变量，例如maxLen ，以存储直到第i个索引为止获得的最长子数组的长度，该索引仅包含偶数或奇数。
- 初始化一个变量，例如Len ，以存储当前子数组直到第i个数组元素的长度，该数组仅由偶数或奇数组成。
- 初始化一个变量，例如MaxElem ，以存储获得的最长子数组的最大元素，直到第i个索引仅由偶数或奇数元素组成。
- 使用变量i遍历数组。对于每个第i个数组元素，检查arr [i]％2是否等于arr [i – 1]％2 。如果发现为真，则增加Len的值。
- 否则，更新Len = 1的值。
- 如果Len> = maxLen ，则更新MaxElem = max(MaxElem，arr [i]) 。
- 最后，输出MaxElem的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the largest element
// of the longest subarray consisting
// only of odd or even elements only
int maxelementLongestSub(int arr[], int n)
{
// Stores largest element of the
// longest subarray till i-th index
int MaxElem = arr[0];
// Stores maximum length of the
// longest subarray till i-th index
int maxLen = 1;
// Stores length of the current
// subarray including the i-th element
int Len = 1;
// Stores largest element in
// current subarray
int Max = arr[0];
// Traverse the array
for (int i = 1; i < n; i++) {
// If arr[i] and arr[i - 1]
// are either even numbers
// or odd numbers
if (arr[i] % 2 == arr[i - 1] % 2) {
// Update Len
Len++;
// Update Max
if (arr[i] > Max)
Max = arr[i];
// If Len greater than
// maxLen
if (Len >= maxLen) {
maxLen = Len;
// Update MaxElem
if (Max >= MaxElem)
MaxElem = Max;
}
}
else {
// Update Len
Len = 1;
// Update Max
Max = arr[i];
// If Len greater
// than maxLen
if (Len >= maxLen) {
// Update maxLen
maxLen = Len;
// If Max greater
// than MaxElem
if (Max >= MaxElem) {
// Update MaxElem
MaxElem = Max;
}
}
}
}
return MaxElem;
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 5, 7, 8, 12, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxelementLongestSub(arr, n);
return 0;
} C
// C program to implement
// the above approach
#include
// Function to find the largest element
// of the longest subarray consisting
// only of odd or even elements only
int maxelementLongestSub(int arr[], int n)
{
// Stores largest element of the
// longest subarray till i-th index
int MaxElem = arr[0];
// Stores maximum length of the
// longest subarray till i-th index
int maxLen = 1;
// Stores length of the current
// subarray including the i-th element
int Len = 1;
// Stores largest element in
// current subarray
int Max = arr[0];
// Traverse the array
for(int i = 1; i < n; i++)
{
// If arr[i] and arr[i - 1]
// are either even numbers
// or odd numbers
if (arr[i] % 2 == arr[i - 1] % 2)
{
// Update Len
Len++;
// Update Max
if (arr[i] > Max)
Max = arr[i];
// If Len greater than
// maxLen
if (Len >= maxLen)
{
maxLen = Len;
// Update MaxElem
if (Max >= MaxElem)
MaxElem = Max;
}
}
else
{
// Update Len
Len = 1;
// Update Max
Max = arr[i];
// If Len greater
// than maxLen
if (Len >= maxLen)
{
// Update maxLen
maxLen = Len;
// If Max greater
// than MaxElem
if (Max >= MaxElem)
{
// Update MaxElem
MaxElem = Max;
}
}
}
}
return MaxElem;
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 5, 7, 8, 12, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", maxelementLongestSub(arr, n));
return 0;
}
// This code is contributed by sourav singh Java
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to find the largest element
// of the longest subarray consisting
// only of odd or even elements only
static int maxelementLongestSub(int arr[], int n)
{
// Stores largest element of the
// longest subarray till i-th index
int MaxElem = arr[0];
// Stores maximum length of the
// longest subarray till i-th index
int maxLen = 1;
// Stores length of the current
// subarray including the i-th element
int Len = 1;
// Stores largest element in
// current subarray
int Max = arr[0];
// Traverse the array
for(int i = 1; i < n; i++)
{
// If arr[i] and arr[i - 1]
// are either even numbers
// or odd numbers
if (arr[i] % 2 == arr[i - 1] % 2)
{
// Update Len
Len++;
// Update Max
if (arr[i] > Max)
Max = arr[i];
// If Len greater than
// maxLen
if (Len >= maxLen)
{
maxLen = Len;
// Update MaxElem
if (Max >= MaxElem)
MaxElem = Max;
}
}
else
{
// Update Len
Len = 1;
// Update Max
Max = arr[i];
// If Len greater
// than maxLen
if (Len >= maxLen)
{
// Update maxLen
maxLen = Len;
// If Max greater
// than MaxElem
if (Max >= MaxElem)
{
// Update MaxElem
MaxElem = Max;
}
}
}
}
return MaxElem;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 3, 5, 7, 8, 12, 10 };
int n = arr.length;
System.out.print(maxelementLongestSub(arr, n));
}
}
// This code is contributed by sourav singhPython3
# Python3 program to implement
# the above approach
# Function to find the largest element
# of the longest subarray consisting
# only of odd or even elements only
def maxelementLongestSub(arr, n):
# Stores largest element of the
# longest sub-array till i-th index
MaxElem = arr[0]
# Stores maximum length of the
# longest sub-array till i-th index
maxLen = 1
# Stores length of the current
# sub-array including the i-th element
Len = 1
# Stores largest element in
# current sub-array
Max = arr[0]
for i in range(1, n):
# If arr[i] and arr[i - 1]
# are either even numbers
# or odd numbers
if arr[i] % 2 == arr[i - 1] % 2:
# Update Len
Len += 1
# Update Max
if arr[i] > Max:
Max = arr[i]
# If Len greater than
# maxLen
if Len >= maxLen:
maxLen = Len
# Update MaxElem
if Max >= MaxElem:
MaxElem = Max
else:
# Update Len
Len = 1
# Update Max
Max = arr[i]
# If Len greater
# than maxLen
if Len >= maxLen:
maxLen = Len
# If Max greater
# than MaxElem
if Max >= MaxElem:
MaxElem = Max
return MaxElem
# Driver Code
arr = [ 1, 3, 5, 7, 8, 12, 10 ]
n = len(arr)
print(maxelementLongestSub(arr, n))
# This code is contributed by sourav singhC#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the largest element
// of the longest subarray consisting
// only of odd or even elements only
static int maxelementLongestSub(int[] arr,
int n)
{
// Stores largest element of the
// longest subarray till i-th index
int MaxElem = arr[0];
// Stores maximum length of the
// longest subarray till i-th index
int maxLen = 1;
// Stores length of the current
// subarray including the i-th element
int Len = 1;
// Stores largest element in
// current subarray
int Max = arr[0];
// Traverse the array
for(int i = 1; i < n; i++)
{
// If arr[i] and arr[i - 1]
// are either even numbers
// or odd numbers
if (arr[i] % 2 == arr[i - 1] % 2)
{
// Update Len
Len++;
// Update Max
if (arr[i] > Max)
Max = arr[i];
// If Len greater than
// maxLen
if (Len >= maxLen)
{
maxLen = Len;
// Update MaxElem
if (Max >= MaxElem)
MaxElem = Max;
}
}
else
{
// Update Len
Len = 1;
// Update Max
Max = arr[i];
// If Len greater
// than maxLen
if (Len >= maxLen)
{
// Update maxLen
maxLen = Len;
// If Max greater
// than MaxElem
if (Max >= MaxElem)
{
// Update MaxElem
MaxElem = Max;
}
}
}
}
return MaxElem;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 3, 5, 7, 8, 12, 10 };
int n = arr.Length;
// Function call
Console.Write(maxelementLongestSub(arr, n));
}
}
// This code is contributed by sourav singhJavascript
输出：
7时间复杂度： O(N)
辅助空间： O(1)