【leetcode】1318. Minimum Flips to Make a OR b Equal to c

题目如下：

Given 3 positives numbers a, b and c. Return the minimum flips required in some bits of a and b to make ( a OR b == c ). (bitwise OR operation).
Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.

Example 1:

Input: a = 2, b = 6, c = 5
Output: 3
Explanation: After flips a = 1 , b = 4 , c = 5 such that (a OR b == c)

Example 2:

Input: a = 4, b = 2, c = 7
Output: 1

Example 3:

Input: a = 1, b = 2, c = 3
Output: 0

Constraints:

• 1 <= a <= 10^9
• 1 <= b <= 10^9
• 1 <= c <= 10^9

解题思路：本题很简单。如果c某一位是0，那么需要a和b相应的那一位都是0，翻转次数为a和b相应位1的个数；如果c的某位是1，那么只有a和b对应位都是0的情况下做两次翻转。

代码如下：

class Solution(object):
    def minFlips(self, a, b, c):
        """
        :type a: int
        :type b: int
        :type c: int
        :rtype: int
        """
        res = 0
        na = bin(a)[2:].rjust(32, '0');
        nb = bin(b)[2:].rjust(32, '0');
        nc = bin(c)[2:].rjust(32, '0');

        for i in range(32):
            ia,ib,ic = int(na[i]),int(nb[i]),int(nc[i])
            if ia | ib == ic:continue
            elif ic == 0:
                res += 1 if ia * ib == 0 else 2
            else:
                res += 1

        return res