Problem 371: Sum of Two Integers

https://leetcode.com/problems/sum-of-two-integers/

思路

https://discuss.leetcode.com/topic/49771/java-simple-easy-understand-solution-with-explanation

https://discuss.leetcode.com/topic/50315/a-summary-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently

这两篇对于 bit manipulation 的总结非常好！

• 首先用 a & b 来得到进位，并且循环的退出条件是什么时候没有进位了，什么时候就退出。

• 然后用 a ^ b 来得到不同的 bit

• 最后把 carrier shift，以此类推

public class Solution {
    public int getSum(int a, int b) {
        if (a == 0) return b;
        if (b == 0) return a;

        while (b != 0) {
            int carrier = a & b;
            a = a ^ b;
            b = (carrier << 1);
        }

        return a;
    }
}