迟来的每日se训练
import java.util.*;
//编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前
class ListNode {
int val;
ListNode next = null;
ListNode(int val) { this.val = val;}}
public class Partition {
public ListNode partition(ListNode pHead, int x) {
// write code here
ListNode bs = null;
ListNode be = null;
ListNode as = null;
ListNode ae = null;
while (pHead != null) {
if(pHead.val < x) {
if(bs == null) {
bs = pHead;
be = pHead;
}else {
be.next = pHead;//此时的cur已经是循环过一次的
be = be.next;//把be和be.next一起往后移
}
}else {
if(as == null) {
as = pHead;
ae = pHead;
}else {
ae.next = pHead;
ae = ae.next;
}
}
pHead = pHead.next;
}
/*
把新的链表串起来
1、两个段内 可能有一个是空的
如果as不等于空，ae.next = null;
2、如果两个段都有数据
be.next = as;
*/
if(bs == null) {
return as;
}
/else(as == null) {
return bs;
}if(bs != null && as != null) {
be.next = as;
ae.next = null;
}
return bs;/
if(bs != null) {
be.next = as;
if (as != null) {
ae.next = as;
}
}
return bs;
}
}
import java.util.Arrays;
public class Gift {
public int getValue(int[] gifts, int n) {
// write code here
Arrays.sort(gifts);
int check = gifts[gifts.length/2];
int count = 0;
for (int i = 0;i < gifts.length;i ) {
if (gifts[i] == check) {
count ;
}
}
if (count > gifts.length/2) {
return check;
}else return 0;
}
public static void main(String[] args) { int[] gifts = {1,5,5,4,4,5,5,}; int n = gifts.length; Gift gift = new Gift(); System.out.println(gift.getValue(gifts,n));}}
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;
public class Main {
//查找和排序 查找和排序 题目:输入任意(用户，成绩)序列，可以获得成绩从高到低或从低到高的排列
// ，相同成绩都按先录入排列在前的规则处理。
private static class Student {
private String name;
private int score;
private Student(String name,int score){ this.name = name; this.score = score; }}public static void main(String[] args) { Scanner sc=new Scanner(System.in); while(sc.hasNext()){ int numPeople=sc.nextInt(); int option=sc.nextInt(); List<Student> list = new ArrayList<>(); for(int i=0;i<numPeople;i ){ list.add(new Student(sc.next(),sc.nextInt())); } if(option==0){ list.sort((o1, o2) -> o2.score - o1.score); }else if(option==1){ list.sort((o1,o2) ->o1.score-o2.score); //list.sort(Comparator.comparingInt(o -> o.score)); } for(Student stu : list){ System.out.println(stu.name " " stu.score); } }}// Scanner sc = new Scanner(System.in);
// while (sc.hasNext()) {
// int n = sc.nextInt();
// int t = sc.nextInt();
// Student[] students = new Student[n];
// for (int i = 0; i < n; i ) {
// String name = sc.nextLine();
// int score = sc.nextInt();
// students[i] = new Student(name, score);
// }
// if (t == 1) {
// for (int i = 0;i < n;i ) {
// for (int j = i 1; j < n;j ) {
// if (students[i].score > students[j].score) {
// Student student1 = new Student(students[j].name,students[j].score);
// students[j] = students[i];
// students[i] = student1;
// }
// }
// }
//
// }
// if (t == 0) {
// for (int i = 0;i < n;i ) {
// for (int j = i 1; j < n;j ) {
// if (students[i].score < students[j].score) {
// Student student1 = new Student(students[i].name,students[i].score);
// students[i] = students[j];
// students[j] = student1;
// }
// }
// }
// }
// StringBuilder sb = new StringBuilder();
// for (Student s : students) {
// sb.append(s.name).append(" ").append(s.score).append(’\n’);
// }
// System.out.print(sb.toString());
// }
}
import java.util.Arrays;
class Substr1 {
//请设计一个高效算法，对于p中的每一个较短字符串，判断其是否为s的子串.
static boolean[] chkSubStr(String[] p, int n, String s) {
boolean[] tf = new boolean[n];
for (int i = 0; i < n; i ) {
tf[i] = s.contains(p[i]);
}
return tf;
}
public static void main(String[] args) { String s = "abcwea"; String[] p = {"a","b","c","q"}; int n = 4; System.out.println(Arrays.toString(chkSubStr(p, n, s)));}}
public class Substr {
//请设计一个高效算法，对于p中的每一个较短字符串，判断其是否为s的子串.
public boolean check(String s1,String s2) {
int i = 0;
while (i s1.length() <= s2.length()-1) {
int a = i;
int count = a s1.length();
StringBuilder sb = new StringBuilder();
sb.append(s2.substring(a,count));
if (sb.toString().equals(s1))
return true;
i ;
}
return false;
}
public boolean[] chkSubStr(String[] p, int n, String s) {
// write code here
boolean[] bn = new boolean[n] ;
for (int i = 0;i < n;i ) {
bn[i] = check(p[i],s);
}
return bn;
}public static void main(String[] args) { Substr substr = new Substr(); String[] p = {"a","b","c"}; int n = 3; String s = "abc"; System.out.println(Arrays.toString(substr.chkSubStr(p, n, s)));}}
import java.util.Scanner;
class dd{
public int foo(int x,int y) {
if (x<=0 || y<=0) {
return 1;
}
return 3*foo(x-6,y/2);
}
}
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
String a = sc.nextLine();
char[] b = a.toCharArray();
for(int i = 0;i< b.length;i ){
char c = b[i];
if(c-‘A’>=0){
c =(char)( c >‘E’? (c-5): (c 21));
b[i] = c;
}
}
}
}
// while (sc.hasNext()) {
// String s = sc.nextLine().toUpperCase();
// char[] ch = s.toCharArray();
// for(int i = 0;i< ch.length;i ) {
// char c = ch[i];
// if (ch[i] < ‘A’) {
// break;
// }
// if (c >= ‘A’) {
// c = (char) (c > ‘E’ ? (c - 5) : (c 21));
// ch[i] = c;
// }
// }
// StringBuilder sb = new StringBuilder();
// sb.append(ch.toString().split(" "));
// System.out.println();
// }
}
//兔子生孩子构成的斐波那契额数列
class ain {
public static int get(int c){
if (c == 1 || c == 2) {
return 1;
}
return get(c-1) get(c-2);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int c = sc.nextInt();
System.out.println(get©);
}
}
}
import java.util.Scanner;
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
//实现一个函数，检查二叉树是否平衡，平衡的定义如下，对于树中的任意一个结点，其两颗子树的高度差不超过1。
//给定指向树根结点的指针TreeNode*root，请返回一个bool，代表这棵树是否平衡.
class Balance {
public boolean isBalance(TreeNode root){
if(root == null){
return true;
}
int left = getTreeHeight(root.left);
int right = getTreeHeight(root.right);
if(Math.abs(left - right) > 1){
return false;
}
return isBalance(root.left) && isBalance(root.right);
}
private int getTreeHeight(TreeNode root) { if(root == null){ return 0; } return Math.max(getTreeHeight(root.left),getTreeHeight(root.right)) 1;}}
//给定-系列正整数，请按要求对数字进行分类，
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()){
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0;i < n;i ){
a[i] = sc.nextInt();
}
int sum0 = 0;int c0 = 0;
int cc = 1;int sum1 = 0;int c1 = 0;
int count = 0;int c2 = 0;
int sum3 = 0;int c3 = 0;
int tmp = 0;int c4 = 0;
for (int i = 0;i < a.length;i ) {
if (a[i]%5 == 0) {
if (a[i]%2 == 0) {
c0 ;
sum0 = a[i];
}
}else if (a[i]%5 == 1) {
c1 ;
sum1 = sum1 cc*a[i];
cc = -1^c1;
}else if (a[i]%5 == 2) {
count ;
c2 ;
}else if (a[i]%5 == 3) {
sum3 =a[i];
c3 ;
}else if (a[i]%5 == 4) {
c4 ;
if (a[i] > tmp) { tmp = a[i]; } } } if (c0 != 0) { System.out.print(sum0 " "); } else { System.out.print("N" " "); } if (c1 != 0) { System.out.print(sum1 " "); } else { System.out.print("N" " "); } if (c2 != 0) { System.out.print(count " "); } else { System.out.print("N" " "); } if (c3 != 0) { System.out.print(1.0*sum3/c3 " "); } else { System.out.print("N" " "); } if (c4 != 0) { System.out.println(tmp); } else { System.out.println("N"); } }}}
import java.util.Scanner;
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
//实现一个函数，检查二叉树是否平衡，平衡的定义如下，对于树中的任意一个结点，其两颗子树的高度差不超过1。
//给定指向树根结点的指针TreeNode*root，请返回一个bool，代表这棵树是否平衡.
class Balance {
public boolean isBalance(TreeNode root){
if(root == null){
return true;
}
int left = getTreeHeight(root.left);
int right = getTreeHeight(root.right);
if(Math.abs(left - right) > 1){
return false;
}
return isBalance(root.left) && isBalance(root.right);
}
private int getTreeHeight(TreeNode root) { if(root == null){ return 0; } return Math.max(getTreeHeight(root.left),getTreeHeight(root.right)) 1;}}
//给定-系列正整数，请按要求对数字进行分类，
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()){
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0;i < n;i ){
a[i] = sc.nextInt();
}
int sum0 = 0;int c0 = 0;
int cc = 1;int sum1 = 0;int c1 = 0;
int count = 0;int c2 = 0;
int sum3 = 0;int c3 = 0;
int tmp = 0;int c4 = 0;
for (int i = 0;i < a.length;i ) {
if (a[i]%5 == 0) {
if (a[i]%2 == 0) {
c0 ;
sum0 = a[i];
}
}else if (a[i]%5 == 1) {
c1 ;
sum1 = sum1 cc*a[i];
cc = -1^c1;
}else if (a[i]%5 == 2) {
count ;
c2 ;
}else if (a[i]%5 == 3) {
sum3 =a[i];
c3 ;
}else if (a[i]%5 == 4) {
c4 ;
if (a[i] > tmp) { tmp = a[i]; } } } if (c0 != 0) { System.out.print(sum0 " "); } else { System.out.print("N" " "); } if (c1 != 0) { System.out.print(sum1 " "); } else { System.out.print("N" " "); } if (c2 != 0) { System.out.print(count " "); } else { System.out.print("N" " "); } if (c3 != 0) { System.out.print(1.0*sum3/c3 " "); } else { System.out.print("N" " "); } if (c4 != 0) { System.out.println(tmp); } else { System.out.println("N"); } }}}