367. Valid Perfect Square
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
Returns: True
Example 2:
Input: 14
Returns: False
题目要求我们在不使用sqrt()的情况下对一个数是否是平方数进行判断，那么有了之前实现sqrt()的经验，我们应该能想到这道题是想考察二分查找了，思路很简单，代码如下:
/*use Binary-Search，0ms*/
class Solution {
public:
bool isPerfectSquare(int num) {
if(num==0)
return false;
long long int low=0,high=num,mid=0;
while(low<=high){
mid=low+((high-low)>>1);
if(mid*mid<num)
low=mid+1;
else if(mid*mid>num)
high=mid-1;
else if(mid*mid==num)
return true;
}
return false;
}
};
牛顿迭代应该也能完成，思路见69. Sqrt(x)的代码。