思路和时间复杂度

1. 思路：利用二分，确定区间是左闭右闭，然后根据大小进行二分
2. 时间复杂度：       

代码

class Solution {
public:
    bool isPerfectSquare(int num) {
        bool flag = false;
        if(num == 0 || num == 1) return true;
        long long left = 0, right = num;
        // 左闭右闭
        while(left <= right){
            long long mid = left + (right - left) / 2;
            if(mid * mid == num){
                return true;
            }else if(mid * mid > num){
                right = mid-1;
            }else{
                left = mid+1;
            }
        }
        return flag;
    }
};