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190 reverse bits

190 reverse bits

作者: larrymusk | 来源:发表于2017-11-20 22:52 被阅读0次

首先取得N比特位的长度,然后把N每个末尾BIT的值保持在VAL中:

uint32_t reverseBits(uint32_t n) {
    int length = sizeof(uint32_t)*8;
    int val = 0;
    while(length--){
        if(n&1)
            val = val << 1|0x1;
        else
            val = val << 1;

        n = n >>1;
    }

    return val;
}

需要理解学习更快的算法:

uint32_t reverseBits(uint32_t n) {
    unsigned int y=0;
    unsigned int bit[32],i;
    for(i=0;i<=31;i++)
    {
        bit[i]=n<<i;
        bit[i]=bit[i]>>31;
        bit[i]=bit[i]<<i;
    }
    for(i=0;i<=31;i++)
    {
       y=y|bit[i];
    }
    return y;
}

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