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Tabulation skills and matrix processing skills
2022-06-12 05:38:00 【Violence produces miracles】
List of articles
Minimum number of bags
Ideas ： First, output a certain number of results in the form of violence , Find the rules according to the results
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class AppleMinBags {
public static int minBags(int apple){
if(apple < 0){
return -1;
}
// Preferential secondary energy installation 8 Start with a bag
int bag8 = apple / 8;
int rest = apple - (bag8 * 8);
while (bag8 >= 0){
if(rest % 6 == 0){
return bag8 + (rest / 6);
}
// The rest cannot be loaded 6 The bag of , Choose to step back
else {
bag8--;
rest += 8;
}
}
return -1;
}
public static void main(String[] args) {
for(int apple = 1; apple < 200;apple++) {
System.out.println(apple + " : "+ minBags(apple));
}
}
}
The law discovered is from 18 Begin to have obvious grouping tendency , Every time 8 In groups , And the odd number returns -1, The remaining results are the current number of groups +3,18 Separate treatment within
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class AppleMinBags {
public static int minBags(int apple){
// Odd returns -1
if((apple & 1) !=0){
return -1;
}
if(apple < 18){
return apple == 0 ? 0 :(apple == 6 || apple == 8) ? 1
: (apple == 12 || apple == 14 || apple == 16) ? 2 : -1;
}
return (apple - 18) / 8 + 3;
}
public static void main(String[] args) {
for(int apple = 1; apple < 200;apple++) {
System.out.println(apple + " : "+ minBags(apple));
}
}
}
The sum of consecutive integers
Ideas ： According to the output of the violence table 200 Whether the number within is the sum of consecutive positive numbers , Try to find out the rules , Then use the rules to write better code
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class MSumToN {
public static boolean isMSum(int num){
for (int i = 1; i <= num; i++) {
int sum = i;
for (int j = i + 1; j <= num; j++) {
if(sum + j > num){
break;
}
if(sum + j == num){
return true;
}
sum += j;
}
}
return false;
}
public static void main(String[] args) {
for (int num = 1; num < 200; num++) {
System.out.println(num + " : " + isMSum(num));
}
}
}
Part of the result 
According to the results, we can find that 2 The number to the power of does not satisfy the condition , So we can get better code according to this rule
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class MSumToN {
public static boolean isMSum(int num){
/* num & (num - 1) Is to judge whether it is 2 Bit operations to the power of because num If it is 2 Power square , Only in binary numbers 1 individual 1 num-1 Will reverse the last few digits ,& The result of the operation will be 0 */
return (num & (num - 1)) != 0;
}
public static void main(String[] args) {
for (int num = 1; num < 200; num++) {
System.out.println(num + " : " + isMSum(num));
}
}
}
Print zigzag matrix
Prepare to point to the starting position in the upper left corner A Point and B spot , Determine row and column boundary conditions ,A Click to move one step to the right first ,B Click to move down one step , Every time AB Form a slash ,A Until you can't go right , Just down ,B Until you can't go down, go right , Finally, control A,B Just don't cross the line , Add a variable to control the printing direction , Every time AB Update variables when slashes are formed , Can do zigzag printing
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class ZigZagPrintMatrix {
public static void printMatrixZigZag(int[][] matrix){
int AR = 0;
int AC = 0;
int BR = 0;
int BC = 0;
int endR = matrix.length - 1;
int endC = matrix[0].length - 1;
boolean fromUp = false;
while(AR != endR +1){
// Print
printLevel(matrix,AR,AC,BR,BC,fromUp);
// After the unified movement
//A Move right until the last column goes down
AR = AC == endC ? AR + 1 : AR;
AC = AC == endC ? AC : AC + 1;
//B Move down until the last line goes right
BC = BR == endR ? BC + 1 : BC;
BR = BR == endR ? BR : BR + 1;
fromUp = !fromUp;
}
System.out.println();
}
public static void printLevel(int[][] m,int AR,int AC,int BR,int BC,boolean f){
// Print from top to bottom
if(f){
while (AR != BR + 1){
System.out.print(m[AR++][AC--] + " ");
}
}
// Print from the bottom up
else{
while (BR != AR - 1){
System.out.print(m[BR--][BC++] + " ");
}
}
}
public static void main(String[] args) {
int[][] matrix = {
{
1, 2, 3, 4 }, {
5, 6, 7, 8 }, {
9, 10, 11, 12 } };
printMatrixZigZag(matrix);
}
}
Rotation print matrix
Ideas ： Prepare variables that mark the upper left column and the lower right column , Print one edge at a time , Start from top edge , Print right first , Until you meet the upper right corner , Just print down , Until you meet the lower right corner , Just print to the left , Until you meet the upper left corner , Be aware of special cases where there is only one row or column , Print by circle , From the outside to the inside
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class PrintMatrixSpiralOrder {
public static void spiralOrderPrint(int[][] matrix){
int tR = 0;
int tC = 0;
int dR = matrix.length - 1;
int dC = matrix[0].length - 1;
while (tR <= dR && tC <= dC){
printEdge(matrix,tR++,tC++,dR--,dC--);
}
}
/** * * @param m * @param tR： Top left row * @param tC： The upper left column * @param dR： The bottom right row * @param dC： The lower right column */
public static void printEdge(int[][] m,int tR,int tC,int dR,int dC){
if(tR == dR){
// Only one line
for (int i = tC; i <= dC; i++) {
System.out.print(m[tR][i] + " ");
}
}
else if(tC == dC){
// There is only one column
for (int i = tR; i <= dR; i++) {
System.out.print(m[i][tC] + " ");
}
}else {
// When the front row and the current row
int curC = tC;
int curR = tR;
while(curC != dC){
System.out.print(m[curR][curC] + " ");
curC++;
}
while (curR != dR){
System.out.print(m[curR][curC] + " ");
curR++;
}
while (curC != tC){
System.out.print(m[curR][curC] + " ");
curC--;
}
while (curR != tR){
System.out.print(m[curR][curC] + " ");
curR--;
}
}
}
public static void main(String[] args) {
int[][] matrix = {
{
1, 2, 3, 4 }, {
5, 6, 7, 8 }, {
9, 10, 11, 12 },
{
13, 14, 15, 16 } };
spiralOrderPrint(matrix);
}
}
Rotate the square matrix clockwise 90 degree
Ideas ： Treat in circles , Rotate from outside to inside , Turn the upper left corner of the circle , Upper right corner , The lower right corner , The lower left corner is classified as 4 A group , And start as an edge in turn , Rotation is to exchange these positions in turn , And through observation , On one side , Take the group on the top side as an example , The number to be exchanged = Top right column position - Top left column position -1, reduce 1 It does not include the upper right corner position , After one revolution, the whole body shrinks to the inner circle , Keep cycling , Until the entire square matrix is rotated
package com.zzf.algorithm;
/** * @author zzf * @date 2022-02-09 */
public class RotateMatrix {
public static void rotate(int[][] matrix){
int a = 0;
int b = 0;
int c = matrix.length - 1;
int d = matrix[0].length - 1;
while(a < c){
rotateEdge(matrix,a++,b++,c--,d--);
}
}
public static void rotateEdge(int[][] m,int a,int b,int c,int d){
int tmp = 0;
for(int i = 0; i< d - b; i++){
tmp = m[a][b+i];
// Lower left corner to upper left corner
m[a][b+i] = m[c - i][b];
// Bottom right corner to bottom left corner
m[c - i][b] = m[c][d - i];
// Top right corner to bottom right corner
m[c][d - i] = m[a + i][d];
// Top left corner to top right corner
m[a + i][d] = tmp;
}
}
public static void printMatrix(int[][] matrix){
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
int[][] matrix = {
{
1, 2, 3, 4 }, {
5, 6, 7, 8 }, {
9, 10, 11, 12 }, {
13, 14, 15, 16 } };
printMatrix(matrix);
rotate(matrix);
System.out.println("=========");
printMatrix(matrix);
}
}
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