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.

以题中例子为例，做一个滤波器在i，j之间：11110000011，来跟N按位与，再把M左移i位，按位或

 class Solution {
     /**
      *@param n, m: Two integer
      *@param i, j: Two bit positions
      *return: An integer
      */
     public int updateBits(int n, int m, int i, int j) {
         // write your code here
         int len = j-i+1;
         int temp = 0;
         for (int x=0; x<len; x++) {
             temp |= 1<<x;
         }
         temp = ~(temp<<i);
         return (n&temp) | (m<<i);
     }
 }