[LintCode] A + B 问题

Bit-by-Bit summation:

 class Solution {
 public:
     /*
      * @param a: The first integer
      * @param b: The second integer
      * @return: The sum of a and b
      */
     int aplusb(int a, int b) {
         // write your code here, try to do it without arithmetic operators.
         int res = , sum = , carry = ;
         for (int i = ; i < ; i++, a >>= , b >>= ){
             int d1 = a & , d2 = b & ;
             sum = (d1 ^ d2 ^ carry);
             carry = max((d1 & d2), max((d1 & carry), (d2 & carry)));
             res ^= (sum << i);
         }
         return res;
     }
 };

Treat a + b as two parts:

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

 class Solution {
 public:
     /*
      * @param a: The first integer
      * @param b: The second integer
      * @return: The sum of a and b
      */
     int aplusb(int a, int b) {
         // write your code here, try to do it without arithmetic operators.
         while (b) {
             int carry = a & b;  // carry
             a ^= b;             // plus without carry
             b = carry << ;     // recursively plus the two parts
         }
         return a;
     }
 };

The code can be further shortened by writing it recursively.

 class Solution {
 public:
     /*
      * @param a: The first integer
      * @param b: The second integer
      * @return: The sum of a and b
      */
     int aplusb(int a, int b) {
         // write your code here, try to do it without arithmetic operators.
         if (!b) return a;
         return aplusb(a ^ b, (a & b) << );
     }
 };

Or just in one line :-)

 class Solution {
 public:
     /*
      * @param a: The first integer
      * @param b: The second integer
      * @return: The sum of a and b
      */
     int aplusb(int a, int b) {
         // write your code here, try to do it without arithmetic operators.
         return (!b ? a : aplusb(a ^ b, (a & b) << ));
     }
 };