Leetcode 190 Reverse Bits

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Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

转换成2进制的string后反转，并且末尾补上原数不足32位的0的个数（乘以2的x次方）。

def reverse_bits(n)
    2**(32-n.to_s(2).length)*n.to_s(2).reverse.to_i(2)
end

 这个方法每次左移一位，末尾增加原来的末位数，循环32次后得到翻转数。

uint32_t reverseBits(uint32_t n) {
    uint32_t m = 0;
    for (int i = 0; i< 32 ; i++,n/=2)
        m = (m<<1) + (n%2);
    return m;
}

Leetcode 190 Reverse Bits

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原文地址：http://www.cnblogs.com/lilixu/p/4645496.html