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2019阿里巴巴编程测试题（在线和手写，记事本写）～～
要求：
给定一个字符串，按照给定的字节数要求输出，
如： str="ABCDEF", 输出4个字节，则为： ABCD.
如： str="哈ABCDEF",输出4个字节，则应为 “哈AB”。
我写的代码是这样的：
@Test
public void test1() { String str="哈abcdf"; System.out.println(outputByByteNum(str,4)); } /** * @Title: outputByByteNum * @Description: 按字节要求输出字符串 * @param: @param str 要求的字符串 * @param: @param num 要求输出的字节数 * @return: String */ public static String outputByByteNum(String str ,int num){ char[] ch=str.toCharArray(); int count=0; StringBuilder sb=new StringBuilder(); for(char c:ch){ if((int)c>255 && count<num) count+=2; else if((int)c<255 && count<num) count+=1; else break; sb.append(c); } return sb.toString(); }希望各位大佬，给一种另外的判断字节数的方式。 谢谢。我这个是根据asc码来判断的。有点牵强
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用非递归实现一个树的前序遍历；
public class Node {
private int num; private Node left; private Node right; public Node Node(){ Node J = new Node(8, null, null); Node H = new Node(4, null, null); Node G = new Node(2, null, null); Node F = new Node(7, null, J); Node E = new Node(5, H, null); Node D = new Node(1, null, G); Node C = new Node(9, F, null); Node B = new Node(3, D, E); Node A = new Node(6, B, C); return A; } public Node(){} public Node(int num, Node left, Node right) { super(); this.num = num; this.left = left; this.right = right; }public int getNum() {
return num; } public Node getLeft() { return left; } public Node getRight() { return right; } @Override public String toString() { return "Node [num=" + num + ", left=" + left + ", right=" + right + "]"; } }
public class Main {
public static void main(String[] args) { Node n=new Node().Node(); posOrder1(n); }//递归前
public static void preFirst(Node A){
if(A!=null){
System.out.println(A);
preFirst(A.getLeft());
preFirst(A.getRight());
}
}
//递归中
public static void midorder(Node A){
if(A!=null){
midorder(A.getLeft());
System.out.println(A);
midorder(A.getRight());
}
}
//递归后
public static void posOrder(Node A){
if(A!=null){
posOrder(A.getLeft());
posOrder(A.getRight());
System.out.println(A);
}
}
//非递归的前序遍历 public static void preOrder1(Node Node) { Stack<Node> stack = new Stack<Node>(); while(Node != null || !stack.empty()) { while(Node != null) { System.out.println(Node + " "); stack.push(Node); Node = Node.getLeft(); } if(!stack.empty()) { Node = stack.pop(); Node = Node.getRight(); } } } //非递归的中序遍历 public static void midOrder1(Node Node) { Stack<Node> stack = new Stack<Node>(); while(Node != null || !stack.empty()) { while(Node != null) { stack.push(Node); Node = Node.getLeft(); } if(!stack.empty()) { Node = stack.pop(); System.out.println(Node + " "); Node = Node.getRight(); } } } //非递归的后序遍历 public static void posOrder1(Node Node) { Stack<Node> stack1 = new Stack<Node>(); Stack<Integer> stack2 = new Stack<>(); int i = 1; while(Node != null || !stack1.empty()) { while (Node != null) { stack1.push(Node); stack2.push(0); Node = Node.getLeft(); }
while(!stack1.empty() && stack2.peek() == i)
{ stack2.pop(); System.out.println(stack1.pop() + " "); }if(!stack1.empty())
{ stack2.pop(); stack2.push(1); Node = stack1.peek(); Node = Node.getRight(); } } } }转载地址：http://wkkin.baihongyu.com/