Given a non-negative integer x, compute and return the square root of x.
Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.
Note: You are not allowed to use any built-in exponent function or operator, such as pow(x, 0.5) or x ** 0.5.
Example 1:
Input: x = 4
Output: 2
Example 2:
Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.
Constraints:
0 <= x <= 231 - 1https://www.youtube.com/watch?v=WjpswYrS2nY&list=PLgUwDviBIf0qYbL4TBaEWgb-ljVdhkM7R&index=3
class Solution {
public:
double multiply(double m,double n)
{
double ans = 1.0;
for(int i=1;i<=n;i++)
ans*=m;
return ans;
}
int mySqrt(int x) {
double n = 2;//Can be generalized for finding the nth root of x
double min = 1;
double max = x;
double epsilon = 1e-6;//number of decimal places we considered for ans is 5
while(max-min>epsilon)
{
double mid = (min+max)/2.0;
if(multiply(mid,n)>x)//target is less than the calculated value
max = mid;
else
min = mid;
}
return (int)max;
}
};