leetcode-190-Reverse Bits

Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

Follow up:

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

将一个数的二进制反过来，求反过来的数。

期，将该数的最低位，作为最高位即可（2^31）,然后依次求解。

例：

对于 0000..........10

ans = 0

ans += 0*(2^31)

ans += 1*(2^30)

...

...

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t i = 1 ,ans = 0,m = 31; // 要用uint32_t 不能用int
        while (m --) { //  将 i = 2^31
            i <<= 1;
        }

        m = 32;
        while (n) {  // m --     用m-- 和n 都可以
            ans += (n&1)*i;
            i >>= 1;
            n >>= 1;
        }
        return ans;
    }
};

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