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.
- code
class Solution {
public int mySqrt(int x) {
if (x < 2) return x;
long num;
int pivot, left = 2, right = x / 2;
while (left <= right) {
pivot = left + (right - left) / 2;
num = (long)pivot * pivot;
if (num > x) right = pivot - 1;
else if (num < x) left = pivot + 1;
else return pivot;
}
return right;
}
}
- code
class Solution:
def mySqrt(self, x: int) -> int:
left, right = 0, x
while True:
mid = left + (right-left)//2
if mid*mid > x:
right = mid - 1
else:
if (mid+1)*(mid+1) > x:
return mid
left = mid + 1