# 367.Valid Perfect Square

**367.Valid Perfect Square**

难度:Easy

> Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

```
Example 1:

Input: 16
Output: true
Example 2:

Input: 14
Output: false
```

不能用sqrt函数，那么就只能直接循环，好在也不需要求出平方根也可以知道大致在什么位置跳出循环。

```
class Solution {
public:
    bool isPerfectSquare(int num) {
        for(int i=1;i<=num;i++)
        {
            long  tmp=i*(long)i;
           if(tmp>num) return false;
            else if (tmp == num ) return true;
        }
        return false;
    }
};
```

这里将i转成long防止整型溢出。当tmp大于num时，即i已经超过了平方根，直接返回false即可。

> 执行用时 :4 ms, 在所有 C++ 提交中击败了81.42%的用户\
> 内存消耗 :8 MB, 在所有 C++ 提交中击败了61.50%的用户


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