367. Valid Perfect Square java solutions

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.

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解法一：

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

解法2：  二分查找法 （java 没有ac 仅供参考）

public class Solution {
    public boolean isPerfectSquare(int num) {
        if(num < 0) return false;
        if(num == 1) return true;
        int low = 0, high = num/2;
        while(low <= high){
            long mid = (low + high)/2;
            long tmp = mid*mid;
            if(tmp == num) return true;
            if(tmp > num) high = mid - 1;
            else low = mid + 1;
        }
        return false;
    }
}

解法3：https://leetcode.com/discuss/110659/o-1-time-c-solution-inspired-by-q_rsqrt   大神的 O(1)  解法。

解法4：

完全平方数是一系列奇数之和，例如：

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
....
1+3+...+(2n-1) = (2n-1 + 1)n/2 = n*n

public class Solution {
    public boolean isPerfectSquare(int num) {
        int i = 1;
        while (num > 0) {
            num -= i;
            i += 2;
        }
        return num == 0;
    }
}