【leetcode】367. Valid Perfect Square

problem

367. Valid Perfect Square

solution：二分法；

class Solution {
public:
    bool isPerfectSquare(int num) {
        int left = 1, right = num;//
        long mid = 0;
        while(left<=right)
        {
            mid = left + 0.5*(right - left);
            long t = mid*mid;
            if(t < num) left = mid+1;
            else if (t > num) right = mid-1;
            else return true;                     
        }
        return false;
    }
};

solution2:

纯数学解法，利用到了这样一条性质，完全平方数是一系列奇数之和;

1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
....
1+3+...+(2n-1) = (2n-1 + 1)n/2 = n*n

时间复杂度为O(sqrt(n))

class Solution {
public:
    bool isPerfectSquare(int num) {
        int i = 1;
        while(num>0)
        {
            num -= i;
            i +=2;
        }
        return num==0;
    }
};

 其他两种方法都出现超时的问题。

 

 

参考

1. Leetcode_367. Valid Perfect Square;

2. GrandYang;

完