69. Sqrt(x)

1. Description

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.

2. Example

Example 1

Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.

Example 2

Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842…, and since we round it down to the nearest integer, 2 is returned.

3. Constraints

• 0 <= x <= 2$^{31}$ - 1

4. Solutions

Binary Search

Time complexity: O(logx)
Space complexity: O(1)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution {
public:
    int mySqrt(int x) {
        int left = 0, right = x;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (mid * mid < x) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return left * left > x ? left - 1 : left;
    }
};