不用四则运算做加法

题目描述

写一个函数，求两个整数之和，要求在函数体内不得使用+、-、*、/四则运算符号。

class Solution {

public:

int Add(int num1, int num2)

{

int sum;

int carry;

while(num2)

{

sum = num1^num2; //换了种形式而已，sum + carry = num1 + num2

carry = (num1&num2)<<1;

num1 = sum;

num2 = carry;

}

return num1;

}

};

**29. Divide Two Integers **

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

代码如下：

class Solution {

public:

int divide(int dividend, int divisor) {

long a = abs((long)dividend);

long b = abs((long)divisor);

long result = 0;

while(a>=b) //减到最后的余数小于b为止

{

long c = b; //一轮一轮的减

int i = 0;

while(a>=c)

{

a = a - c;

result += 1<

i++;

c = c<<1;

}

}

if((dividend>0&&divisor<0)||(dividend<0&&divisor>0))

result = -result;

return (int)result;

}

};

**50. Pow(x, n) **

Implement pow(x, n).

代码如下：

class Solution {

public:

double myPow(double x, int n) {

int flag = 1;

if(x>-0.000001&&x<0.000001)

return 0;

if(n==0)

return 1;

if(n<0)

flag = -1;

if(n%2==0)

{ if(flag>0)

return myPow(x,(flag*n)/2)*myPow(x,(flag*n)/2);

else

return 1/(myPow(x,(flag*n)/2)*myPow(x,(flag*n)/2));

}

else

{

if(flag>0)

return x*myPow(x,((flag*n)-1)/2)*myPow(x,((flag*n)-1)/2);

else

return 1/(x*myPow(x,((flag*n)-1)/2)*myPow(x,((flag*n)-1)/2));

}

}

};

**69. Sqrt(x) **

Implement int sqrt(int x).

Compute and return the square root of x.

class Solution {

public:

int mySqrt(int x) {

if(x<=0)

return 0;

int start = 1;

int end = x;

int mid = 0;

while(start<=end)

{

mid = (start + end)>>1;

if(mid>x/mid)

end = mid - 1;

else if(mid

start = mid + 1;

else

return mid;

}

return end;

}

};