？371. Sum of Two Integers(C++)

最新推荐文章于 2019-05-21 13:21:19 发布

Crystal_ting

最新推荐文章于 2019-05-21 13:21:19 发布

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分类专栏： leetcode

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本文介绍了一种不使用+和-运算符计算两个整数之和的方法，并通过位运算实现。此外，还提供了多种位操作技巧，如设置、清除、测试比特等，并附带示例代码。

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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.

class Solution {
public:
    int getSum(int a, int b) {
        return b == 0 ? a : getSum(a ^ b, (a & b) << 1);
    }
};

https://discuss.leetcode.com/topic/50315/a-summary-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently
补充位运算符相关知识：
Set negation ALL_BITS ^ A or ~A
Set bit A |= 1 << bit
Clear bit A &= ~(1 << bit)
Test bit (A & 1 << bit) != 0
Extract last bit A&-A or A&~(A-1) or x^(x&(x-1))
Remove last bit A&(A-1)
Get all 1-bits ~0
Examples

//Count the number of ones in the binary representation of the given number

int count_one(int n) {
    while(n) {
        n = n&(n-1);
        count++;
    }
    return count;
}

//Is power of four (actually map-checking, iterative and recursive methods can do the same)

bool isPowerOfFour(int n) {
    return !(n&(n-1)) && (n&0x55555555);
    //check the 1-bit location;
}