[LeetCode-190] Reverse Bits（反转位）

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?

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http://blog.csdn.net/xy010902100449/article/details/48241171

【分析】前面有做过反转整数案例，不过这里输入时一个u32 数字，反转之后也不会出现溢出，所以不需要考虑数字溢出的情况。思路如下：通过位移位先将最低位的数字和1相&移动到第31位（最高位），存下来。通过移动输入的u32的数字一位，重新上述操作，和1相&移动第30位（次高位），通过控制 index 即可实现。代码如下：

uint32_t reverseBits(uint32_t n) {    if(0 == n)        return 0;    int index = 32;    uint32_t reverseBitsNum = 0;    while(n) {        reverseBitsNum |= (n&1)<<index; /*高位->低位*/        n = n>>1 /*右移动移位*/        index --;    }    return reverseBitsNum;}