leetcode1318

 1 class Solution:
 2     def converToBin(self,n):
 3         N = [0] * 32
 4         i = 31
 5         while n != 0:
 6             r = n % 2
 7             N[i] = r
 8             i -= 1
 9             n = n // 2
10         return N
11     def minFlips(self, a: int, b: int, c: int) -> int:
12         A,B,C = self.converToBin(a),self.converToBin(b),self.converToBin(c)
13         #print(A,B,C)
14         count = 0
15         for i in range(31,-1,-1):
16             bit_a,bit_b,bit_c = A[i],B[i],C[i]
17             if bit_c == 0:
18                 if bit_a == 1:
19                     count += 1
20                 if bit_b == 1:
21                     count += 1
22             else:#bit_c == 1
23                 if bit_a == 0 and bit_b == 0:
24                     count += 1
25         return count

算法思想：位运算。根据按位与的运算规则：0 | 0 = 0 0 | 1 = 1 1 | 0 = 1 1 | 1 = 1。

先将数字转化为32位的二进制表示。然后对应位置进行比较。

要得到0，需要将所有的1都转化成0；

要得到1，需要至少出现一次1；

posted on 2020-01-12 10:53 Sempron2800+ 阅读(149) 评论(0) 编辑收藏举报