LeetCode | Sqrt (x)

 

Implement int sqrt(int x).

Compute and return the square root of x.

//tag提示用binary search。要求的返回结果是int，不是实际的平方跟，用二分法找到最合适的int就行
public class Solution {
    public int mySqrt(int x) {
        if(x <= 1){
            return x;
        }
        
        int result = 0;
        int left = 1;
        int right = x - 1;                      //在[1~x-1]范围内用二分找结果
        
        while(left <= right){                   //判断时不要用middle*middle == x，
            int middle = (left + right) / 2;    //当middle比较大时，middle*middle可能会溢出越界
            if(middle == x/middle){             //正巧为结果
                result = middle;
                break;
            }
            if(middle < x/middle && (middle+1) > x/(middle+1)){  //近似为结果
                result = middle;
                break;
            }
            
            if(middle < x/middle){
                left = middle + 1;
            }else if(middle > x/middle){
                right = middle - 1;
            }
        }
        
        return result;
    }
}