我试图解决的问题是一个标准的面试问题。给定布尔矩阵，找到从起点到终点的路径。
The start point is assumed the left top corner
The finishing point the right bottom corner.
Only grids with 0 can be moved into.
No diagonal moves are allowed.
这是我的代码。
public class PathFinder {
public static ArrayList<Pair> dfs(int[][] arr, int row, int col, Pair sp, Pair fp){
int[][] check = new int[row][col];
ArrayList<Pair> path = new ArrayList<>();
dfs(arr, row, col, path, check, sp, fp);
return path;
}
private static void dfs(int[][] arr, int row, int col, ArrayList<Pair> path, int[][] check, Pair sp, Pair fp){
if(sp.getRow() == fp.getRow() && sp.getCol() == fp.getCol()) return;
if((sp.getRow() +1 < row) &&(arr[sp.getRow() +1][sp.getCol()] == 0) && (check[sp.getRow()+1][sp.getCol()] == 0)){
check[sp.getRow()+1][sp.getCol()] = 1;
path.add(new Pair(sp.getRow()+1, sp.getCol()));
dfs(arr, row, col, path, check, new Pair(sp.getRow()+1, sp.getCol()), fp);
}else if((sp.getRow() -1 >= 0) &&(arr[sp.getRow() -1][sp.getCol()] == 0) && (check[sp.getRow()-1][sp.getCol()] == 0)){
check[sp.getRow()-1][sp.getCol()] = 1;
path.add(new Pair(sp.getRow()-1, sp.getCol()));
dfs(arr, row, col, path, check, new Pair(sp.getRow()-1, sp.getCol()), fp);
}else if((sp.getCol() +1 < col) &&(arr[sp.getRow()][sp.getCol() +1] == 0) && (check[sp.getRow()][sp.getCol()+1] == 0)){
check[sp.getRow()][sp.getCol()+1] = 1;
path.add(new Pair(sp.getRow(), sp.getCol()+1));
dfs(arr, row, col, path, check, new Pair(sp.getRow(), sp.getCol()+1), fp);
}else if((sp.getCol() -1 >= 0) &&(arr[sp.getRow()][sp.getCol() -1] == 0) && (check[sp.getRow()][sp.getCol()-1] == 0)) {
check[sp.getRow()][sp.getCol() - 1] = 1;
path.add(new Pair(sp.getRow(), sp.getCol() - 1));
dfs(arr, row, col, path, check, new Pair(sp.getRow(), sp.getCol() - 1), fp);
}
}
public static void printPath(ArrayList<Pair> list){
for(Iterator itr = list.iterator(); itr.hasNext();){
Pair p = (Pair) itr.next();
System.out.println(p.getRow()+","+p.getCol());
}
}
}
这是我的配对
public class Pair {
private int row;
private int col;
public Pair(int row, int col){
this.row = row;
this.col = col;
}
public int getRow(){
return row;
}
public int getCol(){
return col;
}
}
这是我的主叫代码。
public class Main {
public static void printArray(int[][] arr, int row, int col){
for (int i = 0; i < row; i++) {
for (int j = 0; j <col ; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
// write your code here
int row = 5;
int col = 7;
int[][] matrix = new int[row][col];
matrix[0][1] = 1;
matrix[0][3] = 1;
matrix[0][5] = 1;
matrix[1][1] = 1;
matrix[1][3] = 1;
matrix[1][6] = 1;
matrix[2][1] = 1;
matrix[2][2] = 1;
matrix[2][6] = 1;
matrix[3][3] = 1;
matrix[3][5] = 1;
matrix[3][6] = 1;
matrix[4][0] = 1;
printArray(matrix, row, col);
ArrayList<Pair> list = PathFinder.dfs(matrix, row, col, new Pair(0,0), new Pair(row-1, col-1));
PathFinder.printPath(list);
}
}
问题是这种深度优先搜索仅适用于特定情况。有人可以帮我修改代码，使其适用于所有情况。请记住，我不想先进行一次呼吸搜索。
答案 0 :(得分：1)
这是一个解决方案，使用包含联结之间的Stack子路径和Pairs的自实现链接列表。已访问的字段保存在矩阵中。最后再次打印矩阵，其中结果字段（找到的路径）的值为3，其他访问的字段的值为2。
public class Pair {
private int row;
private int col;
private Pair next;
public Pair(int row, int col){
this.row = row;
this.col = col;
}
public int getRow(){
return row;
}
public int getCol(){
return col;
}
public Pair getNext() {
return next;
}
public void setNext(Pair next) {
this.next = next;
}
}
///////////////////////
import java.util.*;
public class PathFinder {
private int[][] arr;
private int rowCount;
private int colCount;
private Stack<Pair> junctions = new Stack<>();
public PathFinder(int[][] arr){
this.arr = arr;
this.rowCount = arr.length;
if(rowCount > 0) {
this.colCount = arr[0].length;
}
}
public Pair dfs(Pair sp){
int actualRow = sp.getRow();
int actualCol = sp.getCol();
//we where already here
arr[actualRow][actualCol] = 2;
if(actualRow >= rowCount - 1 && actualCol >= colCount - 1) {
//ready
return sp;
}
boolean deeper = actualRow +1 < rowCount && arr[actualRow +1][actualCol] == 0;
boolean left = actualCol -1 >= 0 && arr[actualRow][actualCol -1] == 0;
boolean right = actualCol +1 < colCount && arr[actualRow][actualCol +1] == 0;
boolean up = actualRow -1 >= 0 && arr[actualRow-1][actualCol] == 0;
//test for junctions
int possibilities = 0;
if(left){
possibilities++;
}
if(right) {
possibilities++;
}
if(deeper){
possibilities++;
}
if(up){
possibilities++;
}
if(possibilities > 1) {
this.junctions.push(sp);
}
Pair nextPair;
if(deeper){
nextPair = new Pair(actualRow + 1, actualCol);
} else if(left) {
nextPair = new Pair(actualRow, actualCol-1);
} else if(right) {
nextPair = new Pair(actualRow, actualCol+1);
} else if(up) {
nextPair = new Pair(actualRow-1, actualCol);
} else {
if(!this.junctions.empty()) {
Pair lastJunction = this.junctions.pop();
lastJunction.setNext(null);
return dfs(lastJunction);
}
return sp;
}
sp.setNext(nextPair);
return dfs(nextPair);
}
}
/////////////////////
public class Main {
public static void printArray(int[][] arr, int row, int col){
for (int i = 0; i < row; i++) {
for (int j = 0; j <col ; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
int rowCount = 6;
int colCount = 8;
int[][] matrix = new int[rowCount][colCount];
matrix[0] = new int[]{0, 1, 0, 1, 0, 0, 0, 1};
matrix[1] = new int[]{0, 1, 0, 0, 0, 1, 0, 0};
matrix[2] = new int[]{0, 0, 0, 1, 0, 1, 0, 0};
matrix[3] = new int[]{0, 1, 1, 1, 1, 0, 0, 1};
matrix[4] = new int[]{0, 0, 0, 0, 0, 1, 0, 0};
matrix[5] = new int[]{0, 1, 0, 1, 0, 0, 1, 0};
printArray(matrix, rowCount, colCount);
Pair pair = new Pair(0,0);
PathFinder finder = new PathFinder(matrix);
Pair finish = finder.dfs(pair);
if(finish.getRow() == rowCount-1 && finish.getCol() == colCount -1) {
while( pair != null){
System.out.println(pair.getRow()+","+pair.getCol());
matrix[pair.getRow()][pair.getCol()] = 3;
pair = pair.getNext();
}
} else {
System.out.println("no path found");
}
printArray(matrix, rowCount, colCount);
}
}