Update Bits详解编程语言

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.

class Solution { 
public: 
    /** 
     [email protected] n, m: Two integer 
     [email protected] i, j: Two bit positions 
     *return: An integer 
     */ 
    int updateBits(int n, int m, int i, int j) { 
        int ones = ~0; 
        int left = ones << (j + 1); 
        int right = ((1 << i) - 1); 
        int mask = left | right; 
 
        return (n & mask) | (m << i); 
    } 
};

class Solution { 
public: 
    /** 
     [email protected] n, m: Two integer 
     [email protected] i, j: Two bit positions 
     *return: An integer 
     */ 
    int updateBits(int n, int m, int i, int j) { 
        int ones = ~0; 
        int mask = 0; 
        if (j < 31) { 
            int left = ones << (j + 1); 
            int right = ((1 << i) - 1); 
            mask = left | right; 
        } else { 
            mask = (1 << i) - 1; 
        } 
 
        return (n & mask) | (m << i); 
    } 
};

class Solution { 
public: 
    /** 
     [email protected] n, m: Two integer 
     [email protected] i, j: Two bit positions 
     *return: An integer 
     */ 
    int updateBits(int n, int m, int i, int j) { 
        // get the bit width of input integer 
        int bitwidth = 8 * sizeof(n); 
        int ones = ~0; 
        // use unsigned for logical shift 
        unsigned int mask = ones << (bitwidth - (j - i + 1)); 
        mask = mask >> (bitwidth - 1 - j); 
 
        return (n & (~mask)) | (m << i); 
    } 
};