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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:
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 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4606409.html
