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

Credits:

Special thanks to @elmirap for adding this problem and creating all test cases.
思路：在binary search time out的情况下，还有另外一种收缩上限的方法，那就是num/i，搜索1到sqrt（num）的上线，这里并没有用sqrt，而是用了除法。很巧妙。
i<=num/i, 实际上i的取值范围是1~ sqrt(num);

public class Solution {
public boolean isPerfectSquare(int num) {
if(num < 0) return false;
if(num == 0 || num == 1) return true;
for(int i=1; i<= num/i; i++){
if( i*i == num) return true;
}
return false;
}
}

Binary Search, timeout.
public class Solution {
public boolean isPerfectSquare(int num) {
if(num < 0) return false;
int start =0; int end = num;
while(start <= end) {
int mid = start+(end-start)/2;
if(mid*mid == num) {
return true;
} else if( mid*mid > num){
end = mid-1;
} else {
start = mid+1;
}
}
return false;
}
}