[LeetCode]#69. Sqrt(x)

Environment: Python 3.7

Key technique: Binary Search, //

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.

Analysis:

Use binary search to find ans. If input is 1, just return answer 1. The brief process is as below.

Solution:

class Solution(object):
    def mySqrt(self, num):
        
        left = 0
        right = num
    
        while(left <= right):
            mid = (left + right)//2
            
            if(mid**2 < num):
                left = mid + 1
                
            elif(mid**2 > num):
                right = mid -1
                
            else:
                return mid
        if num==1:
            return 1
        else:
            return left -1

Submitted result:

Reference:

https://leetcode.com/problems/sqrtx/discuss/557168/python-binary-search-92.59