# Update Bits

## Question

Given two 32-bit numbers, N and M, and two bit positions, i and j.
Write a method to set all bits between i and j in N equal to M
(e g , M becomes a substring of N located at i and starting at j)

Example
Given N=(10000000000)2, M=(10101)2, i=2, j=6

return N=(10001010100)2

Note
In the function, the numbers N and M will given in decimal,
you should also return a decimal number.

Challenge
Minimum number of operations?

Clarification
You can assume that the bits j through i have enough space to fit all of M.
That is, if M=10011，
you can assume that there are at least 5 bits between j and i.
You would not, for example, have j=3 and i=2,
because M could not fully fit between bit 3 and bit 2.

## 题解

Cracking The Coding Interview 上的题，题意简单来讲就是使用 M 代替 N 中的第i位到第j位。很显然，我们需要借用掩码操作。大致步骤如下：

2. 使用mask与 N 进行按位与，清零 N 的第i位到第j位。
3. 对 M 右移i位，将 M 放到 N 中指定的位置。
4. 返回 N | M 按位或的结果。

### C++

class Solution {
public:
/**
*@param n, m: Two integer
*@param i, j: Two bit positions
*return: An integer
*/
int updateBits(int n, int m, int i, int j) {
int ones = ~0;
int left = ones << (j + 1);
int right = ((1 << i) - 1);
int mask = left | right;

return (n & mask) | (m << i);
}
};

### C++

class Solution {
public:
/**
*@param n, m: Two integer
*@param i, j: Two bit positions
*return: An integer
*/
int updateBits(int n, int m, int i, int j) {
int ones = ~0;
if (j < 31) {
int left = ones << (j + 1);
int right = ((1 << i) - 1);
} else {
mask = (1 << i) - 1;
}

return (n & mask) | (m << i);
}
};

### C++

class Solution {
public:
/**
*@param n, m: Two integer
*@param i, j: Two bit positions
*return: An integer
*/
int updateBits(int n, int m, int i, int j) {
// get the bit width of input integer
int bitwidth = 8 * sizeof(n);
int ones = ~0;
// use unsigned for logical shift
unsigned int mask = ones << (bitwidth - (j - i + 1));

return (n & (~mask)) | (m << i);
}
};