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_367_ValidPerfectSquare.java
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package leetcode_1To300;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _367_ValidPerfectSquare {
/**
* 367. Valid Perfect Square
* @param num
* @return
*/
// time : O(sqrt(n)) space : O(1)
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;
}
// time : O(logn) space : O(1)
public boolean isPerfectSquare2(int num) {
int low = 1;
int high = num;
while (low <= high) {
long mid = (high - low) / 2 + low;
if (mid * mid == num) {
return true;
} else if (mid * mid < num) {
low = (int) mid + 1;
} else {
high = (int) mid - 1;
}
}
return false;
}
/**
* 1 = 1
* 4 = 1 + 3
* 9 = 1 + 3 + 5
* 16 = 1 + 3 + 5 + 7
* ...
* @param num
* @return
*/
// time : O(n) space : O(1);
public boolean isPerfectSquare3(int num) {
int i = 1;
while (num > 0) {
num -= i;
i += 2;
}
return num == 0;
}
// Newton Method time : 不知道 space : O(1);
public boolean isPerfectSquare4(int num) {
long x = num;
while (x * x > num) {
x = (x + num / x) / 2;
}
return x * x == num;
}
}