标签：参与 present opera 补码 color 分析 整数 减法 sub
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.
不用+ - 号求两个整数的和，用二进制运算（&, |, ~, ^）
1 public class Solution { 2 public int getSum(int a, int b) { 3 int c = a; 4 int d = b; 5 while ((c & d) != 0){ 6 int cc = c; 7 c = c ^ d; 8 d = (cc & d) << 1; 9 } 10 return (c ^ d); 11 12 } 13 }
"&" AND operation, for example, 2 (0010) & 7 (0111) => 2 (0010)
"^" XOR operation, for example, 2 (0010) ^ 7 (0111) => 5 (0101)
"~" NOT operation, for example, ~2(0010) => -3 (1101) （参与运算的都是以补码的形式）
In bit representation, a = 0001, b = 0011,
First, we can use "and"("&") operation between a and b to find a carry.
carry = a & b, then carry = 0001
Second, we can use "xor" ("^") operation between a and b to find the different bit, and assign it to a,
Then, we shift carry one position left and assign it to b, b = 0010.
Iterate until there is no carry (or b == 0)
1 // Iterative 2 public int getSum(int a, int b) { 3 if (a == 0) return b; 4 if (b == 0) return a; 5 6 while (b != 0) { 7 int carry = a & b; 8 a = a ^ b; 9 b = carry << 1; 10 } 11 12 return a; 13 } 14 15 // Iterative 16 public int getSubtract(int a, int b) { 17 while (b != 0) { 18 int borrow = (~a) & b; 19 a = a ^ b; 20 b = borrow << 1; 21 } 22 23 return a; 24 } 25 26 // Recursive 27 public int getSum(int a, int b) { 28 return (b == 0) ? a : getSum(a ^ b, (a & b) << 1); 29 } 30 31 // Recursive 32 public int getSubtract(int a, int b) { 33 return (b == 0) ? a : getSubtract(a ^ b, (~a & b) << 1); 34 } 35 36 // Get negative number 37 public int negate(int x) { 38 return ~x + 1; 39 }
标签：参与 present opera 补码 color 分析 整数 减法 sub
原文地址：http://www.cnblogs.com/xuehaoyue/p/6360293.html
