1, Fibonacci formula (matrix method, fast power)
/**
* @Author: Gao Yubo
* @Date: 2021/11/27 20:47
*/
public class Fibonacci {
public static void main(String[] args) {
System.out.println(getFibonacci1(20));
System.out.println(getFibonacci2(20));
}
public static int getFibonacci1(int N) {
if (N == 1) {
return 1;
}
if (N == 2) {
return 1;
}
return getFibonacci1(N - 1) + getFibonacci1(N - 2);
}
/**
* Fibonacci formula
* The linear algebraic method is used for calculation
*/
public static int getFibonacci2(int N) {
if (N < 0) {
return -1;
}
if (N == 1 || N == 2) {
return 1;
}
//Get the base matrix of fibonacci
int[][] base = {{1, 1}, {1, 0}};
//Calculate the n-2 power of base
int[][] res = matrixPower(base, N - 2);
return res[0][0] + res[1][0];
}
//Matrix power
public static int[][] matrixPower(int[][] base, int p) {
//Identity matrix
int[][] resMatrix = new int[base.length][base[0].length];
for (int i = 0; i < base.length; i++) {
resMatrix[i][i] = 1;
}
int[][] temp = base;
//Counting
for (; p != 0; p >>= 1) {
if ((p & 1) != 0) {
resMatrix = muliMatrix(resMatrix, temp);
}
temp = muliMatrix(temp, temp);
}
return resMatrix;
}
//Computing matrix multiplication
public static int[][] muliMatrix(int[][] temp, int[][] temp1) {
int [][] res = new int[temp.length][temp1[0].length];
for (int i = 0 ; i < temp.length; i++){
for (int j = 0 ; j < temp1[0].length; j++){
for (int k = 0; k < temp1.length; k++){
res[i][j] += temp[i][k] * temp1[k][j];
}
}
}
return res;
}
}
2, I can't spell a triangle
On the hazy prairie, Xiao Hong picked up n sticks. The length of the i-th stick is I. Xiao Hong is very happy now. I want to choose three sticks to form a beautiful triangle. But Xiao Ming wants to tease Xiao Hong and remove some sticks, so that Xiao Hong can't form a triangle by choosing three sticks at will.
How many sticks should Xiao Ming remove at least?
Given N, return at least how many roots are removed?
/**
* @Author: Gao Yubo
* @Date: 2021/11/27 21:58
*/
public class DeleteWood {
//Make sure the rest are Fibonacci numbers and you can't spell a triangle
public static int minDelete(int m) {
int fiboCount = getFiboCount(m);
return m-fiboCount;
}
/**
* Calculate the number of Fibonacci numbers in the [1-N] number
*/
public static int getFiboCount(int N){
int index1 = 1;
int index2 = 2;
int res = 2;
while (index1 + index2 <= N){
index1 += index2;
index2 = index1 - index2;
res++;
}
return res;
}
public static void main(String[] args) {
int test = 8;
System.out.println(minDelete(test));
}
}
3, 01 Backpack
Niuniu is going to take part in the spring outing organized by the school. Before departure, Niuniu is going to put some snacks into his backpack. Niuniu's backpack has a capacity of w. Niuniu's family has n bags of snacks, and the volume of the I bag of snacks is v[i]. Niuniu wants to know how many kinds of snacks he has when the total volume does not exceed the backpack capacity (a total volume of 0 is also a method).
/**
* @Author: Gao Yubo
* @Date: 2021/11/27 22:12
*/
public class Bag01Problem {
public static void main(String[] args) {
int[] arr = { 4, 3, 2,5,3,5,8,2,3,4,58};
int w = 8;
int[][] ints = new int[arr.length][9];
for (int i = 0 ; i < ints.length ; i++){
Arrays.fill(ints[i],-1);
}
System.out.println(func4(arr, w,0));
System.out.println(func5(arr, w));
}
public static int func4(int[] v,int weight,int index){
if (weight < 0){
return 0;
}
if (weight == 0){
return 1;
}
if (index == v.length-1){
return weight - v[index] >=0?2:1;
}
int s = func4(v,weight-v[index],index+1);
int u = func4(v,weight,index+1);
return s + u;
}
public static int func5(int[]v,int weight){
int[][] dp = new int[v.length][weight + 1];
for (int i = 0; i < v.length; i++){
dp[i][0] = 1;
}
for (int i = 0; i <= weight; i++){
dp[v.length-1][i] = i - v[v.length-1] >=0?2:1;
}
for (int i = v.length-2; i >=0; i--){
for (int j = 0; j <= weight; j++){
dp[i][j] = j-v[i] < 0?dp[i+1][j]:dp[i+1][j-v[i]]+dp[i+1][j];
}
}
return dp[0][weight];
}
}
4, Print directory level
Give you an array arr of string type, for example: String [] arr = {"B \ CST", "d \", "a \ D \ e", "a \ B \ C"}; You draw the directory structure contained in these paths. The subdirectories are directly listed under the parent directory and two spaces to the right than the parent directory, like this:
a
b
c
d
e
b
cst
d
Those at the same level shall be arranged in alphabetical order and shall not be disordered.
/**
* @Author: Gao Yubo
* @Date: 2021/11/28 22:17
* Give you an array arr of string type, for example: String [] arr = {"B \ \ CST", "d \ \", "a \ \ D \ \ e", "a \ \ B \ \ C"};
* You draw the directory structure contained in these paths, and the subdirectories are directly listed under the parent directory and two spaces to the right than the parent directory
*/
public class PrintCatalogue {
public static void main(String[] args) {
String[] arr = { "b\\cst", "d\\", "a\\d\\e", "a\\b\\c" };
print(arr);
}
static class Node{
public String name;
//Using sequential map, you can print in order
public TreeMap<String ,Node> nextMap;
public Node(String name){
this.name = name;
nextMap = new TreeMap<>();
}
}
/**
* Prefix tree mode
* Each letter corresponds to a node. If the to be added is in the next of the current letter, it will be placed in the next. If not, it will be created
*/
public static void print(String[] strings){
Node head = generateFolderTree(strings);
printTree(head,0);
}
private static Node getPrefixTree(String[] strings) {
Node head = new Node("");
Node cur = head;
for (String str: strings){
String[] letters = getLetters(str);
//Update cur back to head
cur = head;
for (String letter: letters){
if (!cur.nextMap.containsKey(letter)) {
cur.nextMap.put(letter,new Node(letter));
}
cur = cur.nextMap.get(letter);
}
}
return head;
}
public static Node generateFolderTree(String[] folderPaths) {
Node head = new Node("");
for (String foldPath : folderPaths) {
String[] paths = foldPath.split("\\\\");
Node cur = head;
for (int i = 0; i < paths.length; i++) {
if (!cur.nextMap.containsKey(paths[i])) {
cur.nextMap.put(paths[i], new Node(paths[i]));
}
cur = cur.nextMap.get(paths[i]);
}
}
return head;
}
public static void printTree(Node head, int level) {
if (level != 0){
System.out.println(getSpace(level)+head.name);
}
for (Node node: head.nextMap.values()){
printTree(node,level+1);
}
}
//Obtain different numbers of spaces according to the number of layers of the tree. There are no spaces in 1 layer and 2 spaces in 2 layers
public static String getSpace(int level){
StringBuilder stringBuilder = new StringBuilder();
for (int i = 1; i < level;i++){
stringBuilder.append(" ");
}
return stringBuilder.toString();
}
//Set "a \ \ B \ \ C" = > [a, B, C]
public static String[] getLetters(String str){
return str.split("\\\\");
}
}
5, Convert bidirectional linked list (binary tree template)
The two-way linked list node structure is the same as the binary tree node structure, if you think last is left and next is next.
Given the head node of a search binary tree, please convert it into an ordered two-way linked list and return the head node of the linked list.
/**
* @Author: Gao Yubo
* @Date: 2021/11/28 22:55
* The two-way linked list node structure is the same as the binary tree node structure, if you think last is left and next is next.
* Given the head node of a search binary tree, please convert it into an ordered two-way linked list and return the chain
* The header node of the table.
*/
public class ConvertDoubleList {
public static void main(String[] args) {
Node head = new Node(5);
head.left = new Node(2);
head.right = new Node(9);
head.left.left = new Node(1);
head.left.right = new Node(3);
head.left.right.right = new Node(4);
head.right.left = new Node(7);
head.right.right = new Node(10);
head.left.left = new Node(1);
head.right.left.left = new Node(6);
head.right.left.right = new Node(8);
Node newHead = getDoubleLinkedList(head);
printDoubleLinkedList(newHead);
}
public static void printDoubleLinkedList(Node head) {
System.out.print("Double Linked List: ");
Node end = null;
while (head != null) {
System.out.print(head.value + " ");
end = head;
head = head.right;
}
System.out.print("| ");
while (end != null) {
System.out.print(end.value + " ");
end = end.left;
}
System.out.println();
}
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
static class Info{
public Node head;
public Node last;
public Info(Node head, Node last) {
this.head = head;
this.last = last;
}
}
public static Node getDoubleLinkedList(Node searchHead){
Info info = f(searchHead);
return info.head;
}
public static Info f(Node node){
if (node == null){
return new Info(null,null);
}
Info leftInfo = f(node.left);
Info rightInfo = f(node.right);
//connect
if (leftInfo.last != null){
leftInfo.last.right = node;
node.left = leftInfo.last;
}
if (rightInfo.head != null){
rightInfo.head.left = node;
node.right = rightInfo.head;
}
Node head = leftInfo.head==null?node:leftInfo.head;
Node last = rightInfo.last==null?node:rightInfo.last;
return new Info(head,last);
}
}
6, Given an integer matrix, returns the maximum cumulative sum of submatrixes.
/**
* @Author: Gao Yubo
* @Date: 2021/11/28 23:37
* Maximum sum of submatrix
*/
public class MaxChildMatrixSum {
public static void main(String[] args) {
int[][] matrix = { { -90, 48, 78 }, { 64, -40, 64 }, { -81, -7, 66 } };
System.out.println(getMaxMatrixSum(matrix));
}
/**
* Add the lower row of the matrix to the upper row, and calculate the cumulative sum of the subarray
*/
public static int getMaxMatrixSum(int[][] matrix){
if(matrix == null || matrix.length == 0){
return 0;
}
//Accumulate the upper layers of the matrix to facilitate the summation of subarrays, -- "change the summation of the matrix into the summation of one-dimensional arrays
int[] sumMatrixArray = null;
int cur = 0;
int max = Integer.MIN_VALUE;
for (int i = 0 ; i < matrix.length; i++){
sumMatrixArray = new int[matrix[0].length];
for (int j = i; j < matrix.length; j++){
//Use maximum subarray and
cur = 0;
for (int k = 0 ; k < matrix[0].length;k++){
//accumulation
sumMatrixArray[k] += matrix[j][k];
cur += sumMatrixArray[k];
max = Math.max(max,cur);
cur = Math.max(0,cur);
}
}
}
return max;
}
}