69. Sqrt(x)
Math, Binary Search, AMateList ·Problem Statement
link: LeetCode.cn LeetCode Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.
You must not use any built-in exponent function or operator.
For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.
Example:
Input: x = 4
Output: 2
Input: x = 8
Output: 2
Constraints: 0 <= x <= 231 - 1
Solution Approach
Use binary search to find the square root by narrowing the range of possible values until the closest integer is found.
Algorithm
- Set Search Range: Start with left at 0 and right at x to cover all possible square root values.
- Binary Search: Repeatedly check the midpoint. If its square is too high, move the right boundary; if too low, move the left boundary and save the midpoint as a potential answer.
- Return Closest: Once the search range is exhausted, return the last midpoint that didn’t exceed x.
Implement
class Solution:
def mySqrt(self, x: int) -> int:
l, r = 0, x
res = 0
while l <= r:
m = l + ((r - l) // 2)
if m ** 2 > x:
r = m - 1
elif m ** 2 < x:
l = m + 1
res = m
else:
return m
return res