069. Sqrt(x)

Question 69

https://leetcode.com/problems/sqrtx/description/

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

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

Answer

v1. digit-by-digit calculation

1. 將上限定位數字本身，下限定為1

2. 開始十分逼近法 https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit-by-digit_calculation

3. 除以二時取整數部分（因題目準確度只要到整數位）

class Solution: def mySqrt(self, x):
    high = x
    low = 1
    
    if x == 0:
        return 0
    # digit-by-digit calculation
    while high - low > 1:
        mid = (low + high)//2
        #print("+++ ",mid)
        if mid**2 > x:
            high = mid
        else:
            low = mid

    return low