完全平方数(Valid Perfect Square)
2019-06-15 本文已影响0人
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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.
简单说就是判断一个整数是否可以表示成另一个数的平方
- 完全平方数可以表示成从1开始的连续奇数的和。
证明: 利用等差数列求和
class Solution {
public:
bool isPerfectSquare(int num)
{
for(int i = 1; num > 0;i += 2)
{
num -= i;
}
return num == 0;
}
};
时间复杂度
- 二分查找法
判断 能否表示为 mid*mid
class Solution {
public:
bool isPerfectSquare(int num)
{
int low = 1, high = num;
long mid;
while (low <= high)
{
mid = low + (high - low) / 2;
long tmp = mid * mid;
if (tmp == num)
return true;
else if (tmp > num)
high = mid - 1;
else
low = mid + 1;
}
return false;
}
};
时间复杂度
- 牛顿迭代法
class Solution {
public:
bool isPerfectSquare(int num)
{
long i = num / 2 + 1;
while (i * i > num) i = (i + num / i) / 2;
return i * i == num;
}
};