Sum of Two Integers，sumoftwointegers

Sum of Two Integers，sumoftwointegers

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.

個人思路：繞開+、-，利用循環和 += 與 -= 。

class Solution {
public:
  int getSum(int a, int b) {
    double x = a, y = b;
    double c = fabs(y);
    for (double i = 1; i <= c; i++) {
      if (b > 0) a++;
      if (b < 0) a--;
    }
    return a;
  }
};

錯誤原因：特殊值的計算。a = 2147483647, b = -2147483647。Time Limit Exceeded

自己在電腦上跑結果是對的，然而循環次數太多了，浪費時間，比較笨的一種方法，年輕人，還是太傻太天真，還需要學習一個。

其他思路：

1.利用下標偏移自帶加法 by dimle

int getSum(int a, int b) {
    auto c = (char*)(a);
    return (int)((int64_t)(&c[b]));
}

把a變成一個地址為a的指針c，把它看成數組，偏移b後，地址變成a + b,再把地址轉成int。（補充學習：<inttypes.h>）

2.利用位運算 by vdvvdd

class Solution {
public:
    int getSum(int a, int b) {
        int sum = a;
        
        while (b != 0)
        {
            sum = a ^ b;//calculate sum of a and b without thinking the carry 
            b = (a & b) << 1;//calculate the carry
            a = sum;//add sum(without carry) and carry
        }
        
        return sum;
    }
};

解釋：

class Solution {
public:
    int getSum(int a, int b) {
        // Take 1 + 2 for example in 4 bits.
        // Same as 0001 + 0010
        // Binary addition requires going through each bit position and
        // generating a carry in bit, carry out bit and sum bit.
        // For example, in bit position 0 in 0001 + 0010, 
        // carry in: 0, sum bit: carry in + 1 + 0 = 1, carry out: 0
        // For bit position 1,
        // carry in: 0, sum bit: carry in + 0 + 1 = 1, carry out: 0
        // Using a truth table, we can figure out that
        // sum bit = carry in xor bit_a xor bit_b.
        // carry out = (carry in AND bit_a) OR (carry in AND bit_b) OR (bit_a AND bit_b)
        // carry out becomes the carry in for the next bit position.
        int result = 0x0;
        int sum_bit = 0x0;
        int carry_bit = 0x0;
        int bitmask = 0x1;
        int bit_a = 0x0;
        int bit_b = 0x0;
        
        int i = 0;  // Keep track of what bit position im in.
        while (bitmask != 0x0) {
            // Isolate bits in each bit position.
            bit_a = (a & bitmask) >> i;
            bit_b = (b & bitmask) >> i;
            
            // Calculate sum, carry_bit is a carry in now.
            sum_bit = ((carry_bit ^ bit_a) ^ bit_b);
            // Shift sum_bit to correct bit position, OR into result.
            result |= (sum_bit << i);
            // Calculate carry out, carry_bit is now a carry out after calculation.
            carry_bit = ((bit_a & bit_b) | (carry_bit & bit_a)) | (carry_bit & bit_b);
            
            // Shift bitmask over 1 to the left.
            bitmask <<= 1;
            // Increment bit position.
            ++i;
        }
        return result;
    }
};

3. 另一種位運算 by cjchan0210

class Solution {
public:
int getSum(int a, int b) {
if(a && b) return getSum(a^b, (a&b) << 1);
else return a|b;
}
};

viewed as binary, c = (a&b)<<1 means carry bits, and d = a^b means sum without carry obviously...
and then get the sum of c and d...
if a or b is 0, return a or b, so that is a|b

拓展閱讀：https://discuss.leetcode.com/topic/50315/a-summary-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently