Valid Perfect Square

Given a positive integernum, write a function which returns True ifnumis a perfect square else False.

Note:Do notuse any built-in library function such assqrt.

Example 1:

Input: 16
Returns: True

Example 2:

Input: 14
Returns: False

Solution

public boolean isPerfectSquare(int num) {
     int i = 1;
     while (num > 0) {
         num -= i;
         i += 2;
     }
     return num == 0;
 }

The time complexity is O(sqrt(n)), a more efficient one using binary search whose time complexity is O(log(n)):

public boolean isPerfectSquare(int num) {
    int low = 1, high = num;
    while (low <= high) {
        long mid = (low + high) >>> 1;
        if (mid * mid == num) {
            return true;
        } else if (mid * mid < num) {
            low = (int) mid + 1;
        } else {
            high = (int) mid - 1;
        }
    }
    return false;
}

One thing to note is that we have to use long for mid to avoid mid * mid from overflow. Also, you can use long type for low and high to avoid type casting for mid from long to int.

And a third way is to use Newton Method to calculate the square root or num, refer to Newton Method for details.

public boolean isPerfectSquare(int num) {
    long x = num;
    while (x * x > num) {
        x = (x + num / x) >> 1;
    }
    return x * x == num;
}