69. Sqrt(x) && 367. Valid Perfect Square

69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

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(M) Pow(x, n) (M) Valid Perfect Square

Time Limit Exceed Solution if start from 0.

public class Solution {
    public int mySqrt(int x) {
        if(x == 0)
            return 0;
        int lowerBound = 1;
        int higherBound = 2;

        //search for a boundary
        while (true) {
          int square = lowerBound * lowerBound;
          if (square == x)
            return lowerBound;
          if (square < x) {
            lowerBound = higherBound;
            higherBound = higherBound * 2;
          } else {
            lowerBound = lowerBound / 2;
            higherBound = higherBound / 2;
            break;
          }
        }

        int left = lowerBound + 1;
        int right = higherBound - 1;
        while (left <= right) {
          int mid = (left + right) / 2;
          int square = mid * mid;
          if (square == x)
            return mid;
          if (square < x)
            left = mid + 1;
          else
            right = mid - 1;
        }
        return left - 1;
    }
}

Another solution if start from both ends of Integer range.

public class Solution {
    public int mySqrt(int x) {
        if (x == 0)
            return 0;
        int left = 1, right = Integer.MAX_VALUE;
        while (left <= right) {
            int mid = left + (right - left)/2;
            int temp = x/mid;//use division because mid*mid may > Integer.max
            if(mid == temp)
                return mid;
            if (mid > temp)
                right = mid - 1;
            else
                left = mid + 1;
        }
        return left-1;
    }
}

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: 16
Returns: True

Example 2:

Input: 14
Returns: False  

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(M) Sqrt(x)

public class Solution {
    public boolean isPerfectSquare(int num) {
        if (num == 0)
            return true;
        int left = 1, right = Integer.MAX_VALUE;
        while (left <= right) {
            int mid = left + (right - left)/2;
            int temp = num/mid;//use division because mid*mid may > Integer.max
            if(mid == temp)
            {
                if(mid*mid == num)
                    return true;
                return false;//Here is the trap! e.g. num = 5, mid = 2
            }
            if (mid > temp)
                right = mid - 1;
            else
                left = mid + 1;
        }
        return false;
    }
}