Easy to understand C++ code (9ms)


  • 2
    F

    Basically, this code is just keeping pop the last bit from n and push it to the end of the return result.

    Do

    • Get last bit from n
    • Push the bit to the end of the result
    • Pop out the last bit of n

    Until n is 0

    • Push remaining 0s to the n by "ret << nShift"
    • Return the result "ret"
            uint32_t ret = 0;
            
            int nShift = 32;
            while (n && nShift--)
            {
                // shift ret to left by one and move a room for the new push
                ret = (ret << 1);
                // Push the last bit of the n to ret
                if (n%2)
                    ret |= 0x1;
                // pop the last element out
                n = (n>>1);
            }
    
            return ret << nShift;

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