贴两道灵活运用位运算知识来解题的

1.

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

class Solution {
public:
int getSum(int a, int b) {
while(b){
int result = a ^ b;
int carry = (a & b)<<1;
a = result,b = carry;
}
return a;
}
};

a,b为整数，a异或(^)b得到的值是不加上进位值的结果，然后(a&b)<<1得到的是所有进位的值，加起来就行了，但是这里加起来还有进位怎么办呢，那就迭代加，如上。

2.

Given two binary strings, return their sum (also a binary string).

For example,

a = “11”

b = “1”

Return “100”.

string addBinary(string a, string b) {
int alen = a.size();
int blen = b.size();
bool carry = false;
string result;
while( alen>0 || blen>0) {
int abit = alen<=0 ? 0 : a[alen-1]-'0';
int bbit = blen<=0 ? 0 : b[blen-1]-'0';
int cbit = carry ? 1 : 0;
result.insert(result.begin(), '0' + ((abit+bbit+cbit) & 1) );// here is the key point
carry = (abit+bbit+cbit>1);
alen--; blen--;
}
if (carry){
result.insert(result.begin(), '1');
}
return result;
}

一位加法器算a+b，用

(a+b)&1

就能得到低位正确的值了，然后carry的值等于

(a+b)>>1

或者

a+b>1

。