最长子数组，使得相邻元素至少有一个公共数字 |套装 – 2
给定一个包含 N 个整数的数组，任务是找到最长子数组的长度，使得子数组的相邻元素至少有一个共同的数字。
例子：
Input : arr[] = {12, 23, 45, 43, 36, 97}
Output : 3
Explanation: The subarray is 45 43 36 which has
4 common in 45, 43 and 3 common in 43, 36.
Input : arr[] = {11, 22, 33, 44, 54, 56, 63}
Output : 4
Explanation: Subarray is 44, 54, 56, 63上一篇文章中讨论的解决方案使用了 O(N) 额外空间。这个问题可以用常数空间来解决。恒定大小的哈希图用于存储给定数组元素中是否存在数字。要检查相邻元素是否有共同的数字，只需要计算两个相邻元素的数字。因此，hashmap 所需的行数可以减少到 2。变量currRow表示当前行， 1 - currRow表示 hashmap 中的前一行。如果相邻元素有共同的数字，则将当前长度增加 1 并将其与最大长度进行比较。否则将当前长度设置为 1。
以下是上述方法的实现：
C++
// CPP program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
#include
using namespace std;
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
int longestSubarray(int arr[], int n)
{
int i, d;
// To mark presence of digit in current
// element.
int hash[2][10];
memset(hash, 0, sizeof(hash));
// To store current row.
int currRow;
// To store maximum length subarray length.
int maxLen = 1;
// To store current subarray length.
int len = 0;
// To store current array element.
int tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0) {
hash[0][tmp % 10] = 1;
tmp /= 10;
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++) {
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow][d] = 0;
// Find all digits in element.
while (tmp > 0) {
hash[currRow][tmp % 10] = 1;
tmp /= 10;
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++) {
if (hash[currRow][d] && hash[1 - currRow][d]) {
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10) {
len = 1;
}
maxLen = max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
int main()
{
int arr[] = { 11, 22, 33, 44, 54, 56, 63 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << longestSubarray(arr, n);
return 0;
} Java
// Java program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
class GFG
{
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
static int longestSubarray(int arr[], int n)
{
int i, d;
// To mark presence of digit in current
// element.
int hash[][] = new int[2][10];
for( i = 0; i < 2; i++)
for(int j = 0; j < 10; j++)
hash[i][j] = 0;
// To store current row.
int currRow;
// To store maximum length subarray length.
int maxLen = 1;
// To store current subarray length.
int len = 0;
// To store current array element.
int tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0)
{
hash[0][tmp % 10] = 1;
tmp /= 10;
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++)
{
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow][d] = 0;
// Find all digits in element.
while (tmp > 0)
{
hash[currRow][tmp % 10] = 1;
tmp /= 10;
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++)
{
if (hash[currRow][d] != 0 && hash[1 - currRow][d] != 0)
{
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10)
{
len = 1;
}
maxLen = Math.max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 11, 22, 33, 44, 54, 56, 63 };
int n = arr.length;
System.out.println( longestSubarray(arr, n));
}
}
// This code is contributed by Arnab KunduPython3
# Python3 program to print the length of the
# longest subarray such that adjacent elements
# of the subarray have at least one digit in
# common
import math
# Function to print the longest subarray
# such that adjacent elements have at least
# one digit in common
def longestSubarray(arr, n):
i = d = 0;
# To mark presence of digit in current
# element.
HASH1 = [[0 for x in range(10)]
for y in range(2)];
# To store current row.
currRow = 0;
# To store maximum length subarray length.
maxLen = 1;
# To store current subarray length.
len1 = 0;
# To store current array element.
tmp = 0;
# Mark the presence of digits
# of first element.
tmp = arr[0];
while (tmp > 0):
HASH1[0][tmp % 10] = 1;
tmp = tmp // 10;
currRow = 1;
# Find digits of each element and check
# if adjacent elements have common digit
# and update len.
for i in range(0, n):
tmp = arr[i];
for d in range(0, 10):
HASH1[currRow][d] = 0;
# Find all digits in element.
while (tmp > 0):
HASH1[currRow][tmp % 10] = 1;
tmp = tmp // 10;
# Find common digit in adjacent element.
for d in range(0, 10):
if (HASH1[currRow][d] and
HASH1[1 - currRow][d]):
len1 += 1;
break;
# If no common digit is found a new subarray
# has to start from current element.
if (d == 10):
len1 = 1;
maxLen = max(maxLen, len1);
currRow = 1 - currRow;
return maxLen;
# Driver Code
arr = [ 11, 22, 33, 44, 54, 56, 63 ];
n = len(arr);
print(longestSubarray(arr, n));
# This code is contributed by chandan_jnuC#
// C# program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
using System;
class GFG
{
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
static int longestSubarray(int []arr, int n)
{
int i, d;
// To mark presence of digit in current
// element.
int[,] hash = new int[2,10];
for( i = 0; i < 2; i++)
for(int j = 0; j < 10; j++)
hash[i,j] = 0;
// To store current row.
int currRow;
// To store maximum length subarray length.
int maxLen = 1;
// To store current subarray length.
int len = 0;
// To store current array element.
int tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0)
{
hash[0,tmp % 10] = 1;
tmp /= 10;
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++)
{
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow,d] = 0;
// Find all digits in element.
while (tmp > 0)
{
hash[currRow,tmp % 10] = 1;
tmp /= 10;
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++)
{
if (hash[currRow,d] != 0 &&
hash[1 - currRow,d] != 0)
{
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10)
{
len = 1;
}
maxLen = Math.Max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
static void Main()
{
int []arr = { 11, 22, 33, 44, 54, 56, 63 };
int n = arr.Length;
Console.WriteLine( longestSubarray(arr, n));
}
}
// This code is contributed by chandan_jnuPHP
0)
{
$hash[0][$tmp % 10] = 1;
$tmp = (int)($tmp / 10);
}
$currRow = 1;
// Find digits of each element and check
// if adjacent elements have common digit
// and update len.
for ($i = 1; $i < $n; $i++)
{
$tmp = $arr[$i];
for ($d = 0; $d <= 9; $d++)
$hash[$currRow][$d] = 0;
// Find all digits in element.
while ($tmp > 0)
{
$hash[$currRow][$tmp % 10] = 1;
$tmp =(int)($tmp/10);
}
// Find common digit in adjacent element.
for ($d = 0; $d <= 9; $d++)
{
if ($hash[$currRow][$d] &&
$hash[1 - $currRow][$d])
{
$len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if ($d == 10)
{
$len = 1;
}
$maxLen = max($maxLen, $len);
$currRow = 1 - $currRow;
}
return $maxLen;
}
// Driver Code
$arr = array( 11, 22, 33, 44, 54, 56, 63 );
$n = count($arr);
echo longestSubarray($arr, $n);
// This code is contributed by chandan_jnu
?>Javascript
输出：
4时间复杂度： O(N)
辅助空间： O(1)