Lintcode: A+B problem - neverlandly - 博客园


Lintcode: A+B problem - neverlandly - 博客园
For given numbers a and b in function aplusb, return the sum of them. Note You don't need to parse the input and output. Just calculate and return. Example If a=1 and b=2 return 3 Challenge Can you do it with out + operation?
考验Bit Operation, 可以用按位^异或两个操作数对应位以及carry,只是carry是1还是0需要分情况讨论。求更优的解法 复制代码 1 class Solution {
位运算实现整数加法本质就是用二进制进行运算。
其主要用了两个基本表达式:
x^y //执行加法,不考虑进位。
(x&y)<<1 //进位操作
令x=x^y ;y=(x&y)<<1 进行迭代,每迭代一次进位操作右面就多一位0,最多需要“加数二进制位长度”次迭代就没有进位了,此时x^y的值就是结果。
我们来做个3位数的加法:
101+011=1000 //正常加法
位运算加法:
(1) 101 ^ 011 = 110
(101 & 011)<<1 = 010
(2) 110 ^ 010 = 100
(110 & 010)<<1 = 100
(3) 100 ^ 100 = 000
(100 & 100)<<1 = 1000
此时进行相加操作就没有进位了,即000 ^ 1000=1000即是最后结果。
   public int aplusb(int a, int b) {
 8         while(b != 0){
 9             int carry = a & b;
10             a = a ^ b;
11             b = carry << 1;
12         }
13         return a;
14     }

   public int aplusb(int a, int b) {
2         // Click submit, you will get Accepted!
3         if (b == 0) return a;
4         int sum = a^b;
5         int carry = (a&b)<<1;
6         return aplusb(sum, carry);
7     }
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