LeetCode | Sqrt (x)

Implement int sqrt(int x).

Compute and return the square root of x.

//tag提示用binary search。要求的返回結果是int，不是實際的平方跟，用二分法找到最合適的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]範圍內用二分找結果
        
        while(left <= right){                   //判斷時不要用middle*middle == x，
            int middle = (left + right) / 2;    //當middle比較大時，middle*middle可能會溢出越界
            if(middle == x/middle){             //正巧爲結果
                result = middle;
                break;
            }
            if(middle < x/middle && (middle+1) > x/(middle+1)){  //近似爲結果
                result = middle;
                break;
            }
            
            if(middle < x/middle){
                left = middle + 1;
            }else if(middle > x/middle){
                right = middle - 1;
            }
        }
        
        return result;
    }
}