Lintcode: A+B problem
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? Clarification Are a and b both 32-bit integers? - Yes.
直接+没什么好说的,关键在于不用+的操作:
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
1 class Solution { 2 /* 3 * param a: The first integer 4 * param b: The second integer 5 * return: The sum of a and b 6 */ 7 public int aplusb(int a, int b) { 8 // Click submit, you will get Accepted! 9 int i = 0; 10 int res = 0; 11 int carry = 0; 12 for (; i<32; i++) { 13 int aa = (a >> i) & 1; 14 int bb = (b >> i) & 1; 15 res |= (aa ^ bb ^ carry) << i; 16 if (aa == 1 && bb == 1 || ((aa ==1 || bb == 1) && carry == 1)) { 17 carry = 1; 18 } 19 else carry = 0; 20 } 21 return res; 22 } 23 };
贴一个别人的简洁思路:
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。
1 class Solution { 2 /* 3 * param a: The first integer 4 * param b: The second integer 5 * return: The sum of a and b 6 */ 7 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 } 15 }
或者用Recursion写:
1 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 }