给定一个由N个整数组成的数组，编写一个程序来打印最长子数组的长度，以使子数组的相邻元素具有至少一位相同的数字。
例子：
Input : 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 : 11 22 33 44 54 56 63
Output : 4
Explanation: Subarray is 44, 54, 56, 63
通常的方法是检查所有可能的子阵列。但是时间复杂度将是O(n 2 )。
一种有效的方法是创建一个hash [n] [10]数组，该数组标记第i个索引号中数字的出现。我们对每个元素进行迭代，并检查相邻元素之间是否有相同的数字。如果他们有一个共同的数字，我们保留长度的计数。如果相邻元素没有相同的数字，我们将计数初始化为零，然后再次开始对子数组进行计数。打印在迭代时获得的最大计数。我们使用哈希数组来最大程度地减少时间复杂度，因为该数量可以在10 ^ 18的范围内，在最坏的情况下将需要进行18次迭代。
下面给出了上述方法的说明：
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 a[], int n)
{
// remembers the occurrence of digits in
// i-th index number
int hash[n][10];
memset(hash, 0, sizeof(hash));
// marks the presence of digit in i-th
// index number
for (int i = 0; i < n; i++) {
int num = a[i];
while (num) {
// marks the digit
hash[i][num % 10] = 1;
num /= 10;
}
}
// counts the longest Subarray
int longest = INT_MIN;
// counts the subarray
int count = 0;
// check for all adjacent elements
for (int i = 0; i < n - 1; i++) {
int j;
for (j = 0; j < 10; j++) {
// if adjacent elements have digit j
// in them count and break as we have
// got at-least one digit
if (hash[i][j] and hash[i + 1][j]) {
count++;
break;
}
}
// if no digits are common
if (j == 10) {
longest = max(longest, count + 1);
count = 0;
}
}
longest = max(longest, count + 1);
// returns the length of the longest subarray
return longest;
}
// Driver Code
int main()
{
int a[] = { 11, 22, 33, 44, 54, 56, 63 };
int n = sizeof(a) / sizeof(a[0]);
// function call
cout << longestSubarray(a, 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 a[], int n) {
// remembers the occurrence of digits in
// i-th index number
int hash[][] = new int[n][10];
// marks the presence of digit in i-th
// index number
for (int i = 0; i < n; i++) {
int num = a[i];
while (num != 0) {
// marks the digit
hash[i][num % 10] = 1;
num /= 10;
}
}
// counts the longest Subarray
int longest = Integer.MIN_VALUE;
// counts the subarray
int count = 0;
// check for all adjacent elements
for (int i = 0; i < n - 1; i++) {
int j;
for (j = 0; j < 10; j++) {
// if adjacent elements have digit j
// in them count and break as we have
// got at-least one digit
if (hash[i][j] == 1 & hash[i + 1][j] == 1) {
count++;
break;
}
}
// if no digits are common
if (j == 10) {
longest = Math.max(longest, count + 1);
count = 0;
}
}
longest = Math.max(longest, count + 1);
// returns the length of the longest subarray
return longest;
}
// Driver Code
public static void main(String[] args) {
int a[] = {11, 22, 33, 44, 54, 56, 63};
int n = a.length;
// function call
System.out.println(longestSubarray(a, n));
}
}
// This code is contributed by 29AjayKumarPython3
# Python 3 program to print the length of the
# longest subarray such that adjacent elements
# of the subarray have at least one digit in
# common.
import sys
# function to print the longest subarray
# such that adjacent elements have at least
# one digit in common
def longestSubarray(a, n):
# remembers the occurrence of digits
# in i-th index number
hash = [[0 for i in range(10)]
for j in range(n)]
# marks the presence of digit in
# i-th index number
for i in range(n):
num = a[i]
while (num):
# marks the digit
hash[i][num % 10] = 1
num = int(num / 10)
# counts the longest Subarray
longest = -sys.maxsize-1
# counts the subarray
count = 0
# check for all adjacent elements
for i in range(n - 1):
for j in range(10):
# if adjacent elements have digit j
# in them count and break as we have
# got at-least one digit
if (hash[i][j] and hash[i + 1][j]):
count += 1
break
# if no digits are common
if (j == 10):
longest = max(longest, count + 1)
count = 0
longest = max(longest, count + 1)
# returns the length of the longest
# subarray
return longest
# Driver Code
if __name__ == '__main__':
a = [11, 22, 33, 44, 54, 56, 63]
n = len(a)
# function call
print(longestSubarray(a, n))
# This code is contributed by
# Sanjit_PrasadC#
// 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;
public class GFG {
// function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
static int longestSubarray(int []a, int n) {
// remembers the occurrence of digits in
// i-th index number
int [,]hash = new int[n,10];
// marks the presence of digit in i-th
// index number
for (int i = 0; i < n; i++) {
int num = a[i];
while (num != 0) {
// marks the digit
hash[i,num % 10] = 1;
num /= 10;
}
}
// counts the longest Subarray
int longest = int.MinValue;
// counts the subarray
int count = 0;
// check for all adjacent elements
for (int i = 0; i < n - 1; i++) {
int j;
for (j = 0; j < 10; j++) {
// if adjacent elements have digit j
// in them count and break as we have
// got at-least one digit
if (hash[i,j] == 1 & hash[i + 1,j] == 1) {
count++;
break;
}
}
// if no digits are common
if (j == 10) {
longest = Math.Max(longest, count + 1);
count = 0;
}
}
longest = Math.Max(longest, count + 1);
// returns the length of the longest subarray
return longest;
}
// Driver Code
public static void Main() {
int []a = {11, 22, 33, 44, 54, 56, 63};
int n = a.Length;
// function call
Console.Write(longestSubarray(a, n));
}
}
// This code is contributed by Rajput-Ji//PHP
输出：
4
时间复杂度： O(n * 10)
最长子数组，以使相邻元素具有至少一个公共数字|套装– 2
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