371. Sum of Two Integers

题目

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.

分析

不用+ - 号求两个整数的和，用二进制运算（&, |, ~, ^）

解答

解法1：（我）（0ms）

public class Solution {
    public int getSum(int a, int b) {
        int c = a;
        int d = b;
        while ((c & d) != 0){
            int cc = c;
            c = c ^ d;
            d = (cc & d) << 1;
        }
        return (c ^ d);

    }
}

解法2+：两种方法（递归、迭代），三种运算（加法、减法、相反数）

>"&" 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) （参与运算的都是以补码的形式）

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) {
    if (a == 0) return b;
    if (b == 0) return a;

    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;
}