[LintCode] A + B 问题

最新推荐文章于 2020-10-22 13:39:16 发布

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最新推荐文章于 2020-10-22 13:39:16 发布

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原文链接：https://yq.aliyun.com/articles/331512

本文介绍了一种不使用算术运算符实现两个整数相加的方法。通过位操作，包括按位异或处理无进位相加及按位与处理进位，并将进位左移一位后递归相加，直至没有进位。

Bit-by-Bit summation:

 1 class Solution {
 2 public:
 3     /*
 4      * @param a: The first integer
 5      * @param b: The second integer
 6      * @return: The sum of a and b
 7      */
 8     int aplusb(int a, int b) {
 9         // write your code here, try to do it without arithmetic operators.
10         int res = 0, sum = 0, carry = 0;
11         for (int i = 0; i < 32; i++, a >>= 1, b >>= 1){
12             int d1 = a & 1, d2 = b & 1;
13             sum = (d1 ^ d2 ^ carry);
14             carry = max((d1 & d2), max((d1 & carry), (d2 & carry)));
15             res ^= (sum << i);
16         }
17         return res;
18     }
19 };

Treat a + b as two parts:

a + b without carry;
carry of a + b;
recursively plus part 1 and part 2 until no carry exists.

 1 class Solution {
 2 public:
 3     /*
 4      * @param a: The first integer
 5      * @param b: The second integer
 6      * @return: The sum of a and b
 7      */
 8     int aplusb(int a, int b) {
 9         // write your code here, try to do it without arithmetic operators.
10         while (b) {
11             int carry = a & b;  // carry
12             a ^= b;             // plus without carry
13             b = carry << 1;     // recursively plus the two parts
14         }
15         return a;
16     }
17 };

The code can be further shortened by writing it recursively.

 1 class Solution {
 2 public:
 3     /*
 4      * @param a: The first integer
 5      * @param b: The second integer
 6      * @return: The sum of a and b
 7      */
 8     int aplusb(int a, int b) {
 9         // write your code here, try to do it without arithmetic operators.
10         if (!b) return a;
11         return aplusb(a ^ b, (a & b) << 1);
12     }
13 };

Or just in one line :-)

 1 class Solution {
 2 public:
 3     /*
 4      * @param a: The first integer
 5      * @param b: The second integer
 6      * @return: The sum of a and b
 7      */
 8     int aplusb(int a, int b) {
 9         // write your code here, try to do it without arithmetic operators.
10         return (!b ? a : aplusb(a ^ b, (a & b) << 1));
11     }
12 };