[19.4] Find the maximum of two numbers WIHTOUT if-else or any other comparison operator

1. Example
int a,b
Math.max(a,b)
Input: 5,10
Output: 10

1. 5<10 X, can not have <
2. 5 -10 =-5 <0 X, can not have <
3. 5 - 10 = -5 = -101 => 010 + 1 = 011 => 1111 1111 .....1011
Negative means the most significant bit is 1
4. 5-10 = -5 ==> 5 - (-5) = 10-> Max
===  =======> a - sigBit*(a-b)
5. E.g., 10,5 ===> 10 - 0*(10-5) = 10
    E.g., 5, 10===> 5  - 1*(5-10) = 10

  2. Implementation

public int getMax(int a , int b)
{
       int c = a-b;
       int sigBit = (c >> 31) & 0x1;
       int max = a - sigBit*c;
       return max;
}

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