两个数字的二进制表示中的最长公共子串
给定两个整数 n 和 m。在数字及其十进制值的二进制表示中找到最长的连续子集。
示例 1：
Input : n = 10, m = 11
Output : 5
Explanation : Binary representation of
10 -> 1010
11 -> 1011
longest common substring in both is 101
and decimal value of 101 is 5示例 2：
Input : n = 8, m = 16
Output : 8
Explanation : Binary representation of
8 -> 1000
16 -> 10000
longest common substring in both is 1000
and decimal value of 1000 is 8示例 3：
Input : n = 0, m = 8
Output : 9
Explanation : Binary representation of
0 -> 0
8 -> 1000
longest common substring in both is 0
and decimal value of 0 is 0问题来源： https://www.geeksforgeeks.org/citrix-interview-experience-set-5-campus/
先决条件：
1) substr C++
2）找到C++
我们将给定的数字转换为它们的二进制表示，并将二进制表示存储在两个字符串中。一旦我们得到字符串，我们通过尝试从最大可能长度开始的所有长度子字符串来找到最长的公共子字符串。
C++
// CPP program to find longest contiguous
// subset in binary representation of given
// two numbers n and m
#include
using namespace std;
// utility function which returns
// decimal value of binary representation
int getDecimal(string s)
{
int len = s.length();
int ans = 0;
int j = 0;
for (int i = len - 1; i >= 0; i--)
{
if (s[i] == '1')
ans += pow(2, j);
j += 1;
}
return ans;
}
// Utility function which convert decimal
// number to its binary representation
string convertToBinary(int n)
{
string temp;
while (n > 0)
{
int rem = n % 2;
temp.push_back(48 + rem);
n = n / 2;
}
reverse(temp.begin(), temp.end());
return temp;
}
// utility function to check all the
// substrings and get the longest substring.
int longestCommon(int n, int m)
{
int mx = -INT_MAX; // maximum length
string s1 = convertToBinary(n);
string s2 = convertToBinary(m);
string res; // final resultant string
int len = s1.length();
int l = len;
// for every substring of s1,
// check if its length is greater than
// previously found string
// and also it is present in string s2
while (len > 0)
{
for (int i = 0; i < l - len + 1; i++)
{
string temp = s1.substr(i, len);
int tlen = temp.length();
if (tlen > mx && s2.find(temp) != string::npos)
{
res = temp;
mx = tlen;
}
}
len = len - 1;
}
// If there is no common string
if (res == "")
return -1;
return getDecimal(res);
}
// driver program
int main()
{
int n = 10, m = 11;
cout << "longest common decimal value : "
<< longestCommon(m, n) << endl;
return 0;
} Java
// Java program to find longest contiguous
// subset in binary representation of given
// two numbers n and m
public class GFG
{
// utility function to check all the
// substrings and get the longest substring.
static int longestCommon(int n, int m)
{
int mx = -Integer.MAX_VALUE; // maximum length
String s1 = Integer.toBinaryString(n);
String s2 = Integer.toBinaryString(m);
String res = null; // final resultant string
int len = s1.length();
int l = len;
// for every substring of s1,
// check if its length is greater than
// previously found string
// and also it is present in string s2
while (len > 0)
{
for (int i = 0; i < l - len + 1; i++)
{
String temp = s1.substring(i, i + len);
int tlen = temp.length();
if (tlen > mx && s2.contains(temp))
{
res = temp;
mx = tlen;
}
}
len = len - 1;
}
// If there is no common string
if(res == "")
return -1;
return Integer.parseInt(res,2);
}
// driver program to test above function
public static void main(String[] args)
{
int n = 10;
int m = 11;
System.out.println("Longest common decimal value : "
+longestCommon(m, n));
}
}
// This code is Contributed by Sumit GhoshPython3
# Python3 program to find longest contiguous
# subset in binary representation of given
# two numbers n and m
# utility function which returns
# decimal value of binary representation
def getDecimal(s):
lenn = len(s)
ans = 0
j = 0
for i in range(lenn - 1, -1, -1):
if (s[i] == '1'):
ans += pow(2, j)
j += 1
return ans
# Utility function which convert decimal
# number to its binary representation
def convertToBinary(n):
return bin(n)[2:]
# utility function to check all the
# substrings and get the longest substring.
def longestCommon(n, m):
mx = -10**9 # maximum length
s1 = convertToBinary(n)
s2 = convertToBinary(m)
#print(s1,s2)
res="" # final resultant string
lenn = len(s1)
l = lenn
# for every subof s1,
# check if its length is greater than
# previously found string
# and also it is present in s2
while (lenn > 0):
for i in range(l - lenn + 1):
temp = s1[i:lenn + 1]
# print(temp)
tlenn = len(temp)
if (tlenn > mx and( s2.find(temp) != -1)):
res = temp
mx = tlenn
lenn = lenn - 1
# If there is no common string
if (res == ""):
return -1
return getDecimal(res)
# Driver Code
n = 10
m = 11
print("longest common decimal value : ",
longestCommon(m, n))
# This code is contributed by Mohit KumarC#
// C# program to find longest contiguous
// subset in binary representation of given
// two numbers n and m
using System;
using System.Collections.Generic;
class GFG
{
// utility function to check all the
// substrings and get the longest substring.
static int longestCommon(int n, int m)
{
int mx = -int.MaxValue; // maximum length
String s1 = Convert.ToString(n, 2);
String s2 = Convert.ToString(m, 2);;
String res = null; // final resultant string
int len = s1.Length;
int l = len;
// for every substring of s1,
// check if its length is greater than
// previously found string
// and also it is present in string s2
while (len > 0)
{
for (int i = 0; i < l - len + 1; i++)
{
String temp = s1.Substring(i, len);
int tlen = temp.Length;
if (tlen > mx && s2.Contains(temp))
{
res = temp;
mx = tlen;
}
}
len = len - 1;
}
// If there is no common string
if(res == "")
return -1;
return Convert.ToInt32(res, 2);
}
// Driver code
public static void Main(String[] args)
{
int n = 10;
int m = 11;
Console.WriteLine("Longest common decimal value : "
+longestCommon(m, n));
}
}
// This code contributed by Rajput-JiJavascript
输出：
longest common decimal value : 5