LeetCode 371 Sum of Two Integers

This problem is a little confusing to me at first. But after I read some articles online, I got to know that it requires bitwise operations.

So basically, we need to use "^" to calculate the sum of two integers, and "&" << 1 to calculate the carry.

Here is recursive way.

public class Solution {
    public int GetSum(int a, int b) {
        if (b == 0) return a;
        
        int sum = a ^ b;
        int carry = (a & b) << 1;
        
        return GetSum(sum, carry);
    }
}

And here is iterative way.

public class Solution {
    public int GetSum(int a, int b) {
        while (b != 0) 
        {
            int carry = (a & b);
            a = a ^ b;
            b = carry << 1;
        }
        return a;
    }
}

注意这个while循环里面，需要我们先计算int carry = (a & b)，不然会报错。

 

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现在用Python来试试，顺便整理一下原来的思路。Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits.但是在Python里面行不通了。从下面的算法可以看出来Python就不太好用了。但是还是记在这里权当是记录了。

下面这个代码来自于这两个博客

http://bookshadow.com/weblog/2016/06/30/leetcode-sum-of-two-integers/

https://www.hrwhisper.me/leetcode-sum-two-integers/

class Solution(object):
    def getSum(self, a, b):
        """
        :type a: int
        :type b: int
        :rtype: int
        """
        MAX_INT = 0x7FFFFFFF
        MIN_INT = 0x80000000
        MASK = 0x100000000
        while b:
            a, b = (a ^ b) % MASK, ((a & b) << 1) % MASK
        return a if a <= MAX_INT else ~((a % MIN_INT) ^ MAX_INT)