367. Valid Perfect Square

Posted 2020-09-04 panini

题目：

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

链接：

https://leetcode.com/problems/valid-perfect-square/#/description

3/12/2017

遇到的问题：

1. num = 5的输入是总是time exceed，原因是mid值不改变，一直在特定值循环。解决方法：引入prevMid

2. 第10，11行的判断，出现的错误是808201，为什么mid值会一直变大呢？原因是如果用mid * mid == num判断的话，左边很有可能溢出。所以改为，商是mid，同时余数为0的判断。注意溢出值！！！

3. 其他解法：用质数的方法来判断，每当遇到能整除的值可以除去，看最后是否=1

 1 public class Solution {
 2     public boolean isPerfectSquare(int num) {
 3         if (num <= 2) return true;
 4         int low = 1, high = num;
 5         int mid = low + (high - low) / 2;
 6         int prevMid = 0;
 7 
 8         while (low <= high && mid != prevMid) {
 9             prevMid = mid;
10             if (num / mid == mid && num % mid == 0) return true;
11             else if (num / mid < mid) {
12                 high = mid;
13                 mid = low + (mid - low) / 2;
14             } else {
15                 low = mid;
16                 mid = mid + (high - mid) / 2;
17             }
18         }
19         return false;
20     }
21 }

看别人算法之后精简，很值得思考的是，为什么low = mid + 1, high = mid -1？

个人猜测，当要改变low/high的时候，mid已经是不正确的值，如果mid + 1/mid - 1是正确的值的话，我们只会在另一个方向做更正，而一旦触及新的刚改变的low/high时，新的mid必定会返回true

 1 public class Solution {
 2     public boolean isPerfectSquare(int num) {
 3         if (num <= 2) return true;
 4         int low = 1, high = num;
 5         int mid;
 6 
 7         while (low <= high) {
 8             mid = low + (high - low) / 2;
 9             if (num / mid == mid && num % mid == 0) return true;
10             else if (num / mid < mid) {
11                 high = mid - 1;
12             } else {
13                 low = mid + 1;
14             }
15         }
16         return false;
17     }
18 }

还有牛顿法，需要记住

1 public boolean isPerfectSquare(int num) {
2         long x = num;
3         while (x * x > num) {
4             x = (x + num / x) >> 1;
5         }
6         return x * x == num;
7     }