javascript - 如何在 O(lg N) 时间内解决完美平方时避免极端情况？

Question

以下代码判断给定数字是否为 O(lg N) 中的完美平方。如何避免硬编码极端情况 if (num === 1) { return true; } 在下面的解决方案中给出？有什么想法吗？

var isPerfectSquare = function(num) {
    let floor = 0, ceiling = num, mid;

    // Corner case
    if (num === 1) {
        return true;
    }

    while (floor != ceiling) {
        mid = floor + (ceiling - floor) / 2 | 0;
        let mul = mid * mid;

        if (mul === num) {
            return true;
        }

        if (mul > num) {
            ceiling = mid;
        }
        else if (mul < num) {
            floor = mid+1;
        }   
    }
    return false;
};

Answer 1

通过保持智能不变量，我们可以更改代码，最终只需要进行一次相等性检查。应该适用于所有输入

var isPerfectSquare = function(num) {
    let floor = 0, ceiling = num+1, mid;
    // We will keep the invariants: floor*floor <= num,
    //   ceiling * ceiling > num

    while (ceiling - floor > 1) {
        mid = floor + (ceiling - floor) / 2 | 0;
        let mul = mid * mid;

        // Move one of floor/ceiling to mid
        // Retains the invariant!
        if (mul > num) {
            ceiling = mid;
        } else {
            floor = mid;
        }   
    }
    return floor * floor === num;
};

我可能搞砸了语言语法，但这个想法应该可行。