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)

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

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.

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

return N=(10001010100)2

Minimum number of operations?

我們把n的第i位到第j位都變成0，然后加上左移i位的m，不就是所求了嗎？
所以把n的第i位到第j位置零，可以這樣操作：
j < 31: 做一個(gè)mask = ~(1 << j+1) - (1 << i)，讓mask成為一個(gè)第i位到第j位都是0，而高于第j位和低于第i位都是1的這樣一個(gè)數(shù)。然后，和n相與，保證了n在j以上高位，i以下低位的數(shù)字不變，且第i到j(luò)位為0.
j >= 31: 做一個(gè)mask = 1 << i - 1，即長(zhǎng)度為i，每一位都為1的數(shù)字。這樣做掩模的原因是，因?yàn)閖超過(guò)了n的最高位，意味著從第i位向上的所有位都會(huì)被m替換，所以只要用這個(gè)掩模和n相與，保留n的最后i位就可以了。
最后與左移i位的數(shù)m相加，返回結(jié)果。

N = (10000000011)2, M = (10101)2, i = 2, j = 6,
mask = ~(1000000 - 100) = ~111100 = 111...111000011 (32 digits),
m << 2 = 1010100, 
mask & n = 11111000011 & 10000000011 = 10000000011
so return 10001010111

class Solution {
    public int updateBits(int n, int m, int i, int j) {
        int mask = 0;
        if (j < 31) mask = ~((1<