Leetcode: Sum of Two Integers && Summary: Bit Manipulation

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

Example:
Given a = 1 and b = 2, return 3.

转自https://discuss.leetcode.com/topic/49771/java-simple-easy-understand-solution-with-explanation/2,注意里面对于减法的讲解

 have been confused about bit manipulation for a very long time. So I decide to do a summary about it here.

"&" AND operation, for example, 2 (0010) & 7 (0111) => 2 (0010)

"^" XOR operation, for example, 2 (0010) ^ 7 (0111) => 5 (0101)

"~" NOT operation, for example, ~2(0010) => -3 (1101) what??? Don't get frustrated here. It's called two's complement.

1111 is -1, in two's complement

1110 is -2, which is ~2 + 1, ~0010 => 1101, 1101 + 1 = 1110 => 2

1101 is -3, which is ~3 + 1

so if you want to get a negative number, you can simply do ~x + 1

Reference:

https://en.wikipedia.org/wiki/Two%27s_complement

https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html

For this, problem, for example, we have a = 1, b = 3,

In bit representation, a = 0001, b = 0011,

First, we can use "and"("&") operation between a and b to find a carry.

carry = a & b, then carry = 0001

Second, we can use "xor" ("^") operation between a and b to find the different bit, and assign it to a,

Then, we shift carry one position left and assign it to b, b = 0010.

Iterate until there is no carry (or b == 0)

 1 // Iterative
 2 public int getSum(int a, int b) { 5 
 6     while (b != 0) {
 7         int carry = a & b;
 8         a = a ^ b;
 9         b = carry << 1;
10     }
11     
12     return a;
13 }
14 
15 // Iterative
16 public int getSubtract(int a, int b) {
17     while (b != 0) {
18         int borrow = (~a) & b;
19         a = a ^ b;
20         b = borrow << 1;
21     }
22     
23     return a;
24 }
25 
26 // Recursive
27 public int getSum(int a, int b) {
28     return (b == 0) ? a : getSum(a ^ b, (a & b) << 1);
29 }
30 
31 // Recursive
32 public int getSubtract(int a, int b) {
33     return (b == 0) ? a : getSubtract(a ^ b, (~a & b) << 1);
34 }
35 
36 // Get negative number
37 public int negate(int x) {
38     return ~x + 1;
39 }
原文地址:https://www.cnblogs.com/EdwardLiu/p/6103856.html