367. Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16Returns: True

Example 2:

Input: 14Returns: False

官方解答：

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 isO(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 avoidmid*mid from overflow. Also, you can use long type forlow 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 toNewton Method for details.

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