1.1.4 Arrays - Efficient Perfect Square Numbers (Leetcode 367)

1. Topic

leetcode link

Given a positive integer num, write a function that returns true if num is a perfect square, and false otherwise.

Advanced: Do not use any built-in library functions such as sqrt.

Example 1:

Input: num = 16
Output: true

Example 2:

Input: num = 14
Output: false

hint:

1 <= num <= 2³¹ - 1

2. Idea

2.1 Violent solution

Because a positive integer num is given, it starts to traverse from 1, and needs to traverse at most $\sqrt{num}+1$ time.

Complexity Analysis

• Time complexity: $O(\sqrt{n})$ , where n is the maximum value of positive integer num. We need to traverse at most$\sqrt{n}+1$ positive integer.
• Space complexity: $O(1)$。

2.2 Binary search

You can use the dichotomy method to search in the left-closed right-closed interval [1, num]

Complexity Analysis

• Time complexity: $O(\log n)$ , where nn is a positive integer$The\; maximum\; value\; of\; num$ .
• Space complexity: $O(1)$

2.3 Newton iteration method

The core idea: with the help of Taylor series, quickly approach the zero point of the function from the initial value.

1. Set up the equation: $y=f(x)=x^2-num$
2. Compute the next iteration value:

• Derivative, $y^{\prime }=2x\_$
• 直线方程， $y-(x_i^2-num)=2x{\_}_i(x-x_i)$
• The intersection of the line equation and the x-axis, $x_{i+1}=x=\frac{1}{2}(x_i+\frac{num}{x_i})$

3. Termination condition: determine whether the difference between the results of two adjacent iterations is less than a very small non-negative number $ϵ$，whereinϵ$ϵ$ can generally take$10^{-6}or10^{-7}$

Complexity Analysis

• Time complexity: $O(\log n)$ , where nn is a positive integer$The\; maximum\; value\; of\; num$ .
• Space complexity: $O(1)$
  

3. Code implementation

binary search

class Solution {
    
    
public:
    bool isPerfectSquare(int num) {
    
    
        int left = 1;
        int right = num;

        while(left <= right){
    
    
            int middle = left + (right - left)/2;
            long square = (long) middle * middle;

            // 小了，找右区间，left右移
            // 大了，找左区间，right左移
            if(square < num) left = middle + 1;
            else if(square > num) right = middle - 1;
            else return true;
        }
        return false;

    }
};

• Note: square has the risk of overflow

Newton iteration method

class Solution {
    
    
public:
    bool isPerfectSquare(int num) {
    
    
        double x0 = num;
        double esp = 1e-6;

        while(true){
    
    
            double x1 = (x0 + num / x0) / 2;
            if((x0 - x1) < esp) break;
            x0 = x1;
        }
        int x = int(x0);
        return x * x == num;
    }
};

• In the Newton iteration method, because $x_i$Always move left, so $x_0-x_1$Must be greater than or equal to zero