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
从1开始加的话会超时。要利用一个数学规律：
1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
so 1+3+...+(2n-1) = (2n-1 + 1)n/2 = nn
复杂度O(sqrt(n))
public boolean isPerfectSquare(int num) {
int i = 1;
while (num > 0) {
num -= i;
i += 2;
}
return num == 0;
}
或者用牛顿法。
https://discuss.leetcode.com/topic/49325/a-square-number-is-1-3-5-7-java-code/2
