<span style="color:#FF0000;">第一步:把输入的数字转为字符串n.ToString() 第二步:求出字符串的长度即为正整数的位数 第三步:从后向前逆序输出</span> 附代码: using System; using System.Collections.Generic; using System.Linq; using System.Text; //给一个正整数, //要求:一.求它是几位数,二.逆序打印出各位数字. namespa
Problem Description A while ago it was quite cumbersome to create a message for the Short Message Service (SMS) on a mobile phone. This was because you only have nine keys and the alphabet has more than nine letters, so most characters could only be
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package a; public class ShuZi { int m; public int getM() { return m; } public void setM(int m) { this.m = m; } public void shu() { System.out.println("输入的数字是:"+m); if(m>99999) { System.out.println("The number is too large"); } else
所谓的BitMap就是用一个bit位来标记某个元素所对应的value,而key即是该元素,由于BitMap使用了bit位来存储数据,因此可以大大节省存储空间. public class Test { //为了方便,假设数据是以数组的形式给我们的 public static Set<Integer> test(int[] arr) { //用来把重复的数返回,存在Set里,这样避免返回重复的数. Set<Integer> output = new HashSet<>();
public class C3 { public static void main(String[] args) { ArrayList<TreeNode> res = generateTrees(5); System.out.println(res); } public static ArrayList<TreeNode> generateTrees(int n) { if(n == 0){ return new ArrayList<TreeNode>(); } re
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example, Given:start ="hit&
题目: Create a method that takes an array/list as an input, and outputs the index at which the sole odd number is located. This method should work with arrays with negative numbers. If there are no odd numbers in the array, then the method should outpu
