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.

 1 class Solution {
 2     /**
 3      *@param n, m: Two integer
 4      *@param i, j: Two bit positions
 5      *return: An integer
 6      */
 7     public int updateBits(int n, int m, int i, int j) {
 8         // write your code here
 9         int len = j-i+1;
10         int temp = 0;
11         for (int x=0; x<len; x++) {
12             temp |= 1<<x;
13         }
14         temp = ~(temp<<i);
15         return (n&temp) | (m<<i);
16     }
17 }