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
题意:判断一个数是否是平方数。
方法一、平方数一定是1+3+5+7+…的和
class Solution {
public:
bool isPerfectSquare(int num) {
if(num<0) return false;
int i=1;
while(num>0){
num-=i;
i+=2;
}
if(num==0) return true;
return false;
}
};
方法二、二分
class Solution {
public:
bool isPerfectSquare(int num) {
if(num<0) return false;
long l=1,r=num;
while(l<=r)
{
long mid=(l+r)/2;
if(mid*mid>num)
r=mid-1;
else if(mid*mid==num)
return true;
else
l=mid+1;
}
return false;
}
};
方法三、牛顿迭代法求开方
https://en.wikipedia.org/wiki/Integer_square_root#Using_only_integer_division
class Solution {
public:
bool isPerfectSquare(int num) {
long x = num;
while(x*x>num)
x=(x+num/x)>>1;
return x*x==num;
}
};