给定一个整数数组，找出包含可以按连续序列排列的数字的最长子数组的长度。
在上一篇文章中，我们讨论了一个假设给定数组中的元素不同的解决方案。在这里，我们讨论一种即使输入数组有重复也能工作的解决方案。
例子：
Input: arr[] = {10, 12, 11};
Output: Length of the longest contiguous subarray is 3
Input: arr[] = {10, 12, 12, 10, 10, 11, 10};
Output: Length of the longest contiguous subarray is 2 这个想法与之前的帖子类似。在上一篇文章中，我们检查了最大值减去最小值是否等于结束索引减去起始索引。由于允许重复元素，我们还需要检查子数组是否包含重复元素。例如，数组 {12, 14, 12} 跟在第一个属性之后，但其中的数字不是连续元素。
为了检查子数组中的重复元素，我们为每个子数组创建一个散列集，如果我们发现一个元素已经在散列中，我们不考虑当前的子数组。
下面是上述想法的实现。
C++
/* CPP program to find length of the largest
subarray which has all contiguous elements */
#include
using namespace std;
// This function prints all distinct elements
int findLength(int arr[], int n)
{
int max_len = 1; // Initialize result
// One by one fix the starting points
for (int i=0; i myset;
myset.insert(arr[i]);
// Initialize max and min in
// current subarray
int mn = arr[i], mx = arr[i];
// One by one fix ending points
for (int j=i+1; j Java
/* Java program to find length of the largest subarray which has
all contiguous elements */
import java.util.*;
class Main
{
// This function prints all distinct elements
static int findLength(int arr[])
{
int n = arr.length;
int max_len = 1; // Initialize result
// One by one fix the starting points
for (int i=0; i set = new HashSet<>();
set.add(arr[i]);
// Initialize max and min in current subarray
int mn = arr[i], mx = arr[i];
// One by one fix ending points
for (int j=i+1; j Python3
# python program to find length of the largest
# subarray which has all contiguous elements */
# This function prints all distinct elements
def findLenght(arr, n):
max_len = 1
for i in range(0,n - 1):
# Create an empty hash set and
# add i'th element to it
myset = set()
myset.add(arr[i])
# Initialize max and min in
# current subarray
mn = arr[i]
mx = arr[i]
for j in range(i + 1,n):
# If current element is already
# in hash set, then this subarray
# cannot contain contiguous elements
if arr[j] in myset:
break
# Else add current element to hash
# set and update min, max if required.
myset.add(arr[j])
mn = min(mn, arr[j])
mx = max(mx, arr[j])
# We have already checked for
# duplicates, now check for other
#property and update max_len
# if needed
if mx - mn == j - i:
max_len = max(max_len, mx - mn + 1)
return max_len # Return result
# Driver code
arr = [10, 12, 12, 10, 10, 11, 10]
n = len(arr)
print("Length of the longest contiguous subarray is",
findLenght(arr,n))
# This code is contributed by Shrikant13C#
/* C# program to find length of the largest
subarray which has all contiguous elements */
using System;
using System.Collections.Generic;
class GFG {
// This function prints all distinct
// elements
static int findLength(int []arr)
{
int n = arr.Length;
int max_len = 1; // Initialize result
// One by one fix the starting points
for (int i = 0; i < n-1; i++)
{
// Create an empty hash set and
// add i'th element to it.
HashSet set = new HashSet();
set.Add(arr[i]);
// Initialize max and min in current
// subarray
int mn = arr[i], mx = arr[i];
// One by one fix ending points
for (int j = i+1; j < n; j++)
{
// If current element is already
// in hash set, then this subarray
// cannot contain contiguous
// elements
if (set.Contains(arr[j]))
break;
// Else add curremt element to
// hash set and update min,
// max if required.
set.Add(arr[j]);
mn = Math.Min(mn, arr[j]);
mx = Math.Max(mx, arr[j]);
// We have already checked for
// duplicates, now check for
// other property and update
// max_len if needed
if (mx-mn == j-i)
max_len = Math.Max(max_len,
mx - mn + 1);
}
}
return max_len; // Return result
}
// Driver function
public static void Main()
{
int []arr = {10, 12, 12, 10, 10, 11, 10};
Console.WriteLine("Length of the longest"
+ " contiguous subarray is " +
findLength(arr));
}
}
// This code is contributed by Sam007 Javascript
输出：
Length of the longest contiguous subarray is 2如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。