我们给了两个数字n和m，以及两个位置i和j。从j到i开始将m的位插入n。我们可以假设从j到i的位有足够的空间来容纳所有m。也就是说，如果m = 10011，则可以假定j和i之间至少有5位。例如，您不会拥有j = 3和i = 2，因为m不能完全适合第3位和第2位。
例子 ：
Input : n = 1024
m = 19
i = 2
j = 6;
Output : n = 1100
Binary representations of input numbers
m in binary is (10011)2
n in binary is (10000000000)2
Binary representations of output number
(10000000000)2
Input : n = 5
m = 3
i = 1
j = 2
Output : 7
算法 ：
1. Clear the bits j through i in n
2. Shift m so that it lines up with bits j through i
3. Return Bitwise AND of m and n.
最棘手的部分是步骤1。如何清除n中的位?我们可以戴着口罩做到这一点。该掩码将全为1，但j到i位中的0除外。我们先创建遮罩的左半部分，然后再创建右半部分，以创建此遮罩。
以下是上述方法的实现。
C++
// C++ program for implementation of updateBits()
#include
using namespace std;
// Function to updateBits M insert to N.
int updateBits(int n, int m, int i, int j)
{
/* Create a mask to clear bits i through j
in n. EXAMPLE: i = 2, j = 4. Result
should be 11100011. For simplicity, we'll
use just 8 bits for the example. */
int allOnes = ~0; // will equal sequence of all ls
// ls before position j, then 0s. left = 11100000
int left= allOnes << (j + 1);
// l's after position i. right = 00000011
int right = ((1 << i) - 1);
// All ls, except for 0s between i and j. mask 11100011
int mask = left | right;
/* Clear bits j through i then put min there */
int n_cleared = n & mask; // Clear bits j through i.
int m_shifted = m << i; // Move m into correct position.
return (n_cleared | m_shifted); // OR them, and we're done!
}
// Driver Code
int main()
{
int n = 1024; // in Binary N= 10000000000
int m = 19; // in Binary M= 10011
int i = 2, j = 6;
cout << updateBits(n,m,i,j);
return 0;
} Java
// Java program for implementation of updateBits()
class UpdateBits
{
// Function to updateBits M insert to N.
static int updateBits(int n, int m, int i, int j)
{
/* Create a mask to clear bits i through j
in n. EXAMPLE: i = 2, j = 4. Result
should be 11100011. For simplicity, we'll
use just 8 bits for the example. */
int allOnes = ~0; // will equal sequence of all ls
// ls before position j, then 0s. left = 11100000
int left= allOnes << (j + 1);
// l's after position i. right = 00000011
int right = ((1 << i) - 1);
// All ls, except for 0s between i and j. mask 11100011
int mask = left | right;
/* Clear bits j through i then put min there */
// Clear bits j through i.
int n_cleared = n & mask;
// Move m into correct position.
int m_shifted = m << i;
// OR them, and we're done!
return (n_cleared | m_shifted);
}
public static void main (String[] args)
{
// in Binary N= 10000000000
int n = 1024;
// in Binary M= 10011
int m = 19;
int i = 2, j = 6;
System.out.println(updateBits(n,m,i,j));
}
}Python3
# Python3 program for implementation
# of updateBits()
# Function to updateBits M insert to N.
def updateBits(n, m, i, j):
# Create a mask to clear bits i through
# j in n. EXAMPLE: i = 2, j = 4. Result
# should be 11100011. For simplicity,
# we'll use just 8 bits for the example.
# will equal sequence of all ls
allOnes = ~0
# ls before position j,
# then 0s. left = 11100000
left = allOnes << (j + 1)
# l's after position i. right = 00000011
right = ((1 << i) - 1)
# All ls, except for 0s between
# i and j. mask 11100011
mask = left | right
# Clear bits j through i then put min there
n_cleared = n & mask
# Move m into correct position.
m_shifted = m << i
return (n_cleared | m_shifted)
# Driver Code
n = 1024 # in Binary N = 10000000000
m = 19 # in Binary M = 10011
i = 2; j = 6
print(updateBits(n, m, i, j))
# This code is contributed by Anant Agarwal.C#
// C# program for implementation of
// updateBits()
using System;
class GFG {
// Function to updateBits M
// insert to N.
static int updateBits(int n, int m,
int i, int j)
{
/* Create a mask to clear bits i
through j in n. EXAMPLE: i = 2,
j = 4. Result should be 11100011.
For simplicity, we'll use just 8
bits for the example. */
// will equal sequence of all ls
int allOnes = ~0;
// ls before position j, then 0s.
// left = 11100000
int left= allOnes << (j + 1);
// l's after position i.
// right = 00000011
int right = ((1 << i) - 1);
// All ls, except for 0s between i
// and j. mask 11100011
int mask = left | right;
/* Clear bits j through i then put
min there */
// Clear bits j through i.
int n_cleared = n & mask;
// Move m into correct position.
int m_shifted = m << i;
// OR them, and we're done!
return (n_cleared | m_shifted);
}
public static void Main()
{
// in Binary N= 10000000000
int n = 1024;
// in Binary M= 10011
int m = 19;
int i = 2, j = 6;
Console.WriteLine(updateBits(n, m, i, j));
}
}
//This code is contributed by Anant Agarwal.PHP
输出 ：
1100 // in Binary (10001001100)2