Sqrt(x)

应该找一个平方比x小的数，end比start大，如果end的平方就满足标准 那么start肯定满足标准，所以先check end

犯错：最后check应该先判断x/ start >= start，因为找一个平方比x小的最大值

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

class Solution {
    public int mySqrt(int x) {
        if(x == 0)
            return 0;
        
        
        //binary search
        
        int start = 1, end = x;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            
            if(x / mid == mid){
                return mid;
            }
            
            else if(x / mid > mid)
                start = mid;
            else
                end = mid;
        }
        
        //应该先check end        
        if(x/ start >= start){
            return start;
        }
        
        else 
            return end;
        
    }
}

class Solution {
    public int mySqrt(int x) {
        if(x < 2) return x;
        
        
        int start = 0, end = x;

        while(start <= end){
            int mid = start+(end - start) /2;
           
            if( x / mid == mid ) 
                return mid;
            
            else if(x / mid < mid) end = mid -1 ;
            
            else
                start = mid +1;
        }
        
        return end;
       
    }
}