【leetcode】190. Reverse Bits(Python & C++)
2017-09-15 20:29
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190. Reverse Bits
题目链接190.1 题目描述:
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?
190.2 解题思路:
思路一:参考于十进制转二进制,我们手动计算时,都是每次除以2,取余数,然后将所有余数翻转即为二进制数,既然要翻转其二进制,那再转换二进制时,就不用翻转余数了,直接利用余数求出新的翻转后的十进制就可以。例如:10,二进制是1010,10%2=0,5%2=1,2%2=0,1%2=1,所以是倒叙1010,然后在取余的过程中,直接从2的31次方开始算起。
0*2^31+1*2^30+0*29+1*2^28即为翻转二进制后的十进制数。
思路二:利于对bit的按位与操作和左移、右移操作。以8bit举例:
abcdefgh->efghabcd->ghefcdab->hgfedcba,最终完成翻转。
用到32位上,就是先16位交换,再8位、4位、2位、1位。
n = (n >> 16) | (n << 16); n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4); n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2); n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
以上面代码第二行为例,设n=ABCD,其中ABCD各代表8bit数。
则n & 0xff00ff00相当于只保存了A0C0,右移8位后,变为0A0C,
则n & 0x00ff00ff变为0B0D,左移8位后,变为B0D0,
两者按位与后,变为BADC,完成了8位交换。
所以整个过程就是,n=AB,变为n=BA,
n=ABCD,变为n=BADC,
n=ABCDEFGH,变为n=CDABGHEF,
n=ABCDEFGHIJKLMNOP,变为n=BADCFEHGJILKNMPO,
然后是32位,26个字母表示不下了,不写了。
190.3 C++代码:
1、思路一代码(3ms):class Solution122 { public: uint32_t reverseBits(uint32_t n) { uint32_t r=0; vector<int>v; while (n>0) { v.push_back(n % 2); n = n / 2; } for (int i = 0; i < v.size();i++) { r = r + pow(2, 31 - i)*v[i]; } return r; } };
2、思路二代码(3ms)
class Solution122_1 {
public:
uint32_t reverseBits(uint32_t n) {
n = (n >> 16) | (n << 16); n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4); n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2); n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
return n;
}
};
190.4 Python代码:
1、思路一代码(49ms)import math class Solution: # @param n, an integer # @return an integer def reverseBits(self, n): v=[] while n>0: v.append(n%2) n=n/2 a=0 for i in range(len(v)): a=a+math.pow(2,31-i)*v[i] return int(a)
2、思路二代码(52ms)
class Solution1: # @param n, an integer # @return an integer def reverseBits(self, n): n=(n>>16)|(n<<16) n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8) n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4) n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2) n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1) return n
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