每天一道算法题【蓝桥杯】【x的平方根】

思路

使用二分查找模型来避免超时

条件为

mid * mid <= x
mid*mid>x

注意使用longlong类型

#define _CRT_SECURE_NO_WARNINGS 1

class Solution {
public:
    int mySqrt(int x) {
        long long left = 0, right = x;//定义成longlong类型防止内存溢出
        long long mid = 0;
        while (left < right)
        {
            mid = left + (right - left + 1) / 2;  //朴素的二分查找模型
            if (mid * mid <= x)  left = mid; //二分查找的两段性
            else right = mid - 1;
        }
        return left;
    }
}