LintCode Update Bits

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)

Have you met this question in a real interview? Yes

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.

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) {
// write your code here
int mask = ((1<<i) - 1);
mask |= (j == 31) ? 0 : ~((1<<(j + 1)) - 1);
int res = (mask & n) | (m << i);
return res;
}
};