leetcode-69-sqrt(x)

class Solution {
public:
    int (int x) {
    	if(x == 0)
    		return 0;
  		int e = x / 2 + 1;
  		int s = 1;   
  		while(s <= e){
  			//防止s+e越界，所以使用此公式求解中间值
  			int mid = s + (e-s)/2;
  			long long sq = (long long)mid * (long long)mid;
  			if(sq == x) return mid;
  			if(sq < x)
  				s = mid + 1;
  			else e = mid - 1;
  		}
  		return e;   
    }
};

//方法二、牛顿迭代公式