02 第一种解法 暴力解法,定义一个long类型的变量,for循环判断其平方是否小于num,但是不要写循环体,这样省时,最后直接判断其平方是否等于num。 public boolean isPerfectSquare(intnum) {if(num<=1) {returnnum==1; } long i =1;for( ; i*i <num; i++);returni*i ==num; } 0
classSolution {publicbooleanisRectangleCover(int[][] rectangles) {intlx = Integer.MAX_VALUE, ly = Integer.MAX_VALUE, rx = 0, ry = 0, sum = 0; HashSet<String> set =newHashSet<String>();for(int[] rec : rectangles) { lx= Math.min(rec[0], lx); ly= Math.min(rec[1], ly);...
https://leetcode.com/problems/valid-perfect-square/ 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 Ret...
public boolean isPerfectSquare(int num) { if (num < 1) return false; for (int i = 1; num > 0; i += 2) num -= i; return num == 0; } 除了上面的两种方法,我们还可以使用牛顿法,具体牛顿法,大家可以去看一些博客或者看wiki,主要就是切线逼近的思想,将 x2=n 转化为 x2−n=0 的方...
leetcode 279. Perfect Squares Given a positive integern, find the least number of perfect square numbers (for example,1, 4, 9, 16, ...) which sum ton. Example 1: Input:n=12Output: 3 Explanation:12 = 4 + 4 + 4. 1. 1.
public class HuiWen { public static void main(String[] args) { Scanner s = new Scanner(System.in); System.out.print("请输入一个正整数:"); long a = s.nextLong(); String ss = Long.toString(a); char[] ch = ss.toCharArray(); boolean is = true; int j = ch.length; for (int ...
For extra points, reuse local variable names inside {} blocks. The goal is to force the maintenance programmer to carefully examine the scope of every instance. In particular, in Java, make ordinary methods masquerade as constructors. Åccented Letters Use accented characters on variable names. ...
Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two. For example: "112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8. 代码语言:javascript 代码运行次数:0 运行 复制 1 + 1 ...
367 367. Valid Perfect Square.java Easy [Binary Search, Math] O(logN) O(1) Java 456 270 270. Closest Binary Search Tree Value.java Easy [BST, Binary Search, Tree] O(logn) O(1) Java 457 28 28. Implement strStr().java Easy [String, Two Pointers] Java 458 1106 1106. Parsing A ...
sqrt(n))+1)] dp = [float('inf')] * (n+1) # bottom case dp[0] = 0 for i in range(1, n+1): for square in square_nums: if i < square: break dp[i] = min(dp[i], dp[i-square] + 1) return dp[-1] Java 实现...