LeetCode OJ：Sqrt(x)（平方根）

Implement int sqrt(int x).

Compute and return the square root of x.

简单的二分法，注意mid应该选为long，否则容易溢出：

 class Solution {
 public:
     int mySqrt(int x) {
         if(x ==  || x == ) return x;
         int beg = ;
         int end = x;
         long mid = ;   //这里用long，否则会溢出
         while(beg <= end){
             mid = beg + (end - beg)/;
             if(mid * mid < x){
                 beg = mid + ;
             }else if(mid * mid > x){
                 end = mid - ;
             }else{
                 return mid;
             }
         }
         return end;
     }
 };