leetcode每日一题-367:有效的完全平方数

这个其实是很简单的浮点数二分问题,我们只需要用浮点数二分来求出sqrt(num)然后再进行判断即可

C++

class Solution {
public:
    bool isPerfectSquare(int num) {
        double l = 0, r = num;
        while(fabs(r - l) > 1e-6)
        {
            double mid = (l + r) / 2;
            if(mid * mid >= num) r = mid;
            else l = mid;
        }

        int x = (int)r;

        return x * x == num;
    }
};

java

官方题解采用的是整数二分

class Solution {
    public boolean isPerfectSquare(int num) {
        int left = 0, right = num;
        while (left <= right) {
            int mid = (right - left) / 2 + left;
            long square = (long) mid * mid;
            if (square < num) {
                left = mid + 1;
            } else if (square > num) {
                right = mid - 1;
            } else {
                return true;
            }
        }
        return false;
    }
}

作者：LeetCode-Solution