题目描述
Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Note:
- The given integer is guaranteed to fit within the range of a 32-bit signed integer.
- You could assume no leading zero bit in the integer’s binary representation.
Example 1:
Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010\. So you need to output 2.
Example 2:
Input: 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0\. So you need to output 0.
解法一
思路
这题最容易想到的方法就是直接用~
按位取反了,但这样有个明显的问题就是即使是前缀的0也会被取反,如int型5的二进制表示是0000 0000 0000 0000 0000 0000 0000 0101
,其按位取反的结果是1111 1111 1111 1111 1111 1111 1111 1010
,这样前面就多了29个1,我们还得想办法把这些1变成0,从而得到0000 0000 0000 0000 0000 0000 0000 0010
,即十进制2。
但我们不妨换个思路,用异或^
来求解,比如101 ^ 111 = 010
。那么怎么得到111
呢?考虑111 + 1 = 1000
,而1000
又是 最小的 大于101
的 只有一位是1 的二进制数。
所以解决方法出来了:
- 找到最小的大于原数字的二进制值仅有一位为1的数;
- 将此数减1;
- 与原数字按位求异或。
Python
class Solution(object):
def findComplement(self, num):
"""
:type num: int
:rtype: int
"""
all_ones = 1
while all_ones <= num:
all_ones <<= 1
all_ones -= 1
return all_ones ^ num
Java
public class Solution {
public int findComplement(int num) {
long allOnes = 1;
while (allOnes <= num) {
allOnes <<= 1;
}
allOnes -= 1;
return (int) allOnes ^ num;
}
}
C++
class Solution {
public:
int findComplement(int num) {
long allOnes = 1;
while (allOnes <= num) {
allOnes <<= 1;
}
allOnes -= 1;
return (int) allOnes ^ num;
}
};
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