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)

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.

这题如果转化为二进制字符串再操作不难。
但如何用位操作来达到目标比较难。

• 如果j<31，即j不在最高位上。可以把i到j位清为0，可以((1<<(j+1))-(1<<i))得到i到j之间全是1的数，再取反，得到i到j之间全是0的数。
• 如果j=32，(1<<(j+1))即(1<<32)，相当于1<<1 不可行。可以直接(1<<i)-1 得到i到j之间全是0,其他地方是1的数。
• 上面得到的数成为掩码
• (m<<i)+(n&mask) 可以得到最终解。

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;        if(j<31){            mask=~((1<<(j+1))-(1<<i));        }else{            mask=(1<<i)-1;        }        return (m<<i)+(n&mask);    }};