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
代码
Approach #1 Binary Search
class Solution {
public boolean isPerfectSquare(int num) {
if (num < 2) return true;
long left = 2, right = num / 2;
int x, guessSquared;
while (left <= right) {
x = left + (right - left) / 2;
guessSquared = x * x;
if (guessSquared == num) {
return true;
}
if (guessSquared > num) {
right = x - 1;
} else {
left = x + 1;
}
}
return false;
}
}
Approach #2 Newton's Method
Time O(logN) Space O(1)
class Solution {
public boolean isPerfectSquare(int num) {
if (nums < 2) return true;
long x = num / 2;
while (x * x > num) {
x = (x + num / x) / 2;
}
return (x * x == num);
}
}