leetcode367- Valid Perfect Square- easy

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

算法：1.二分法O(logN)。用mid*mid与num比较。细节是看到这种乘法在中间的都要小心溢出，mid用long格式！！不然如果mid比较大100000乘一下溢出变成负数那你就不能得到正确的比较结果了。

2.数学法(O(sqrtN))。因为所有的完美平方数都是1+3+5+...得到的，那你就可以这样加下去到逼近它。

1.二分法(O(logN))

class Solution {
    public boolean isPerfectSquare(int num) {
        if (num < 0) {
            return false;
        }
        long start = 1;
        long end = num;
        while (start + 1 < end) {
            long mid = start + (end - start) / 2;
            if (mid * mid < num) {
                start = (int) mid;
            } else {
                end = (int) mid;
            } 
        }
        if (start * start == num || end * end ==  num) {
            return true;
        }
        return false;
    }
}

2.O(sqrtN)

class Solution {
    public boolean isPerfectSquare(int num) {

        int sum = 0;
        int digit = 1;
        while (sum < num) {
            sum += digit;
            digit += 2;
        }
        if (sum == num) {
            return true;
        } 
        return false;
    }
}