Update Bits

Description:

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.

Solution:

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) { return n & ~((j == 31 ? 0xffffffff : (1<<j+1)-1) - (1<<i)+1) | m<<i; } };

---

Steps:

(e.g. n = (100,0000,0000)2, m = (1,0101)2, i = 2, j = 6)

(1<<j+1) - (1<<i) : 0000,0000,0000,0000,0000,0000,0111,1100~((1<<j+1) - (1<<i)) : 1111,1111,1111,1111,1111,1111,1000,0011n & ~((1<<j+1) - (1<<i)) : 0000,0000,0000,0000,0000,0100,0000,0000m<<i : 101,0100n & ~((1<<j+1) - (1<<i)) | m<<i : 0000,0000,0000,0000,0000,0100,0101,0100

---

转载于:https://www.cnblogs.com/deofly/p/update-bits.html