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)
// Iterative
public int getSum(int a, int b) {
while (b != 0) {
int carry = a & b;
a = a ^ b;
b = carry << 1;
} return a;
} // Iterative
public int getSubtract(int a, int b) {
while (b != 0) {
int borrow = (~a) & b;
a = a ^ b;
b = borrow << 1;
} return a;
} // Recursive
public int getSum(int a, int b) {
return (b == 0) ? a : getSum(a ^ b, (a & b) << 1);
} // Recursive
public int getSubtract(int a, int b) {
return (b == 0) ? a : getSubtract(a ^ b, (~a & b) << 1);
} // Get negative number
public int negate(int x) {
return ~x + 1;
}
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