367. Valid Perfect Square

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

方法一、源自于数学方法，时间复杂度为O（sqrt(num)）
证明：
1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
......
1+3+.....(2n-1)=n/2*(2n-1+1)=n*n
代码：
 bool isPerfectSquare(int num) {
        //时间复杂度为O(sprt(num))
        int i=1;
        while(num>0)
        {
            num-=i;
            i+=2;
        }
        
        return num==0;
    }

方法二、二分查找法，时间复杂度为：O(lgn)
代码:
bool isPerfectSquare(int num) {
        //时间复杂度为O(lg(num))
        int low = 1;
        int high = num;
        while(low <= high)
        {
            long mid = (low + high) / 2;
            if(mid * mid == num)
                return true;
            else if(mid * mid < num)
                low = (int)mid + 1;
            else
                high = (int)mid - 1;
        }
        
        return false;
    }
方法三、牛顿法
牛顿法详细描述见：http://blog.youkuaiyun.com/wangxiaojun911/article/details/18203333
bool isPerfectSquare(int num) {
        long x = num;
        while (x * x > num) {
            x = (x + num / x) >> 1;
        }
        return x * x == num;
    }