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class Solution | ||
{ | ||
public int findComplement(int a) | ||
{ | ||
String s=""; | ||
while(a>0) | ||
{ | ||
if(a%2==1) | ||
{ | ||
s+=""+0; | ||
} | ||
else | ||
{ | ||
s+=""+1; | ||
} | ||
a=a/2; | ||
} | ||
s=new StringBuffer(s).reverse().toString(); | ||
return Integer.parseInt(s,2); | ||
} | ||
} |
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Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation. | ||
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||
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. |