[LintCode]A+B問題

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 };