動機

我忘了lowerbound

Problem

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 = 4Output: 2

Example 2:

Input: x = 8Output: 2Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.

 

Constraints:

  • 0 <= x <= 231 - 1

Sol

所以我直接做bsearch，但要處理loop，loop發生在left等於mid時

class Solution:
    def mySqrt(self, x: int) -> int:
        a,b = 0,x+1 # 猜數字
        while a<b:
            mid = (a+b)//2
            tmp = mid**2
            if tmp == x or mid == a:
                return mid
            elif tmp > x:
                b = mid
            else:
                a = mid
        return a