490. The Maze
原题链接：https://leetcode.com/articles/the-maze/
这道题目是需要冲会员才能使用的，然而我个穷逼现在还是失业状态根本冲不起。。。以后如果把免费题目都刷完了的话，再来冲会员刷这些题目吧！
我的思路
迷宫类问题首先想到的就是回溯法了，思考+实现用了近 3 个小时，终于又写出了简单粗暴的实现来了：
import java.util.Stack;
/**
* Created by clearbug on 2018/2/26.
*/
public class Solution {
public static void main(String[] args) {
Solution s = new Solution();
/**
* 0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0
*/
int[][] board = {
{0, 0, 1, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 1, 0},
{1, 1, 0, 1, 1},
{0, 0, 0, 0, 0},
};
System.out.println(s.traverse(board, 0, 4, 4, 4));
/**
* 0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0
*/
int[][] board2 = {
{0, 0, 1, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 1, 0},
{1, 1, 0, 1, 1},
{0, 0, 0, 0, 0},
};
System.out.println(s.traverse(board2, 0, 4, 3, 2));
}
public Stack<String> traverse(int[][] board, int rowStart, int colStart, int rowDest, int colDest) {
board[rowStart][colStart] = -1; // 标记起始位置已来过
Stack<String> res = new Stack<>();
boolean dfsRes = dfs(board, rowStart, colStart, rowDest, colDest, res);
System.out.println(dfsRes);
return res;
}
public boolean dfs(int[][] board, int rowStart, int colStart, int rowDest, int colDest, Stack<String> res) {
if (rowStart == rowDest && colStart == colDest) { // 说明已抵达目的地
return true;
}
boolean exist = false;
// up
int upRow = rowStart - 1;
while (upRow >= 0 && board[upRow][colStart] < 1) { // board[upRow][colStart] = 0 or board[upRow][colStart] = -1
exist = true;
upRow--;
}
if (exist) {
upRow++;
res.push("up");
if (upRow == rowDest && colStart == colDest) { // 说明已抵达目的地
return true;
}
if (board[upRow][colStart] == -1) { // 说明来过这个位置，目前又循环回来了，所以是死路
res.pop();
} else {
board[upRow][colStart] = -1; // 标记这个位置已来过
if (dfs(board, upRow, colStart, rowDest, colDest, res)) {
return true;
} else {
res.pop();
}
}
}
exist = false;
// down
int downRow = rowStart + 1;
while (downRow < board.length && board[downRow][colStart] < 1) { // board[downRow][colStart] = 0 or board[downRow][colStart] = -1
exist = true;
downRow++;
}
if (exist) {
downRow--;
res.push("down");
if (downRow == rowDest && colStart == colDest) { // 说明已抵达目的地
return true;
}
if (board[downRow][colStart] == -1) { // 说明来过这个位置，目前又循环回来了，所以是死路
res.pop();
} else {
board[downRow][colStart] = -1; // 标记这个位置已来过
if (dfs(board, downRow, colStart, rowDest, colDest, res)) {
return true;
} else {
res.pop();
}
}
}
exist = false;
// left
int leftCol = colStart - 1;
while (leftCol >= 0 && board[rowStart][leftCol] < 1) {
exist = true;
leftCol--;
}
if (exist) {
leftCol++;
res.push("left");
if (rowStart == rowDest && leftCol == colDest) {
return true;
}
if (board[rowStart][leftCol] == -1) {
res.pop();
} else {
board[rowStart][leftCol] = -1;
if (dfs(board, rowStart, leftCol, rowDest, colDest, res)) {
return true;
} else {
res.pop();
}
}
}
exist = false;
// right
int rightCol = colStart + 1;
while (rightCol < board[rowStart].length && board[rowStart][rightCol] < 1) {
exist = true;
rightCol++;
}
if (exist) {
rightCol--;
res.push("right");
if (rowStart == rowDest && rightCol == colDest) {
return true;
}
if (board[rowStart][rightCol] == -1) {
res.pop();
} else {
board[rowStart][rightCol] = -1;
if (dfs(board, rowStart, rightCol, rowDest, colDest, res)) {
return true;
} else {
res.pop();
}
}
}
return false;
}
}
当然了，可以使用类似骑士游历问题中的预测算法来提高效率。下面就来看看官方解答是怎么的吧！
官方方法一（深入优先搜索）
深度优先搜索，英文全称为 Depth First Search，简称为 DFS。毫无疑问，跟我的思路是一样的，就是代码更加简洁，所以我的解法其实就是深度优先搜索啦！
官方解法二（广度优先搜索）
广度优先搜索，英文全称 Breadth First Search，简称 BFS。不过我思考了下，这种广度优先搜索算法最终是无法求出小球所走过的路径的，只能判断小球最终是否会到达目的地。还是按照惯例抄袭一遍代码就当学会了吧：
import java.util.LinkedList;
import java.util.Stack;
import java.util.Queue;
/**
* Created by clearbug on 2018/2/26.
*/
public class Solution {
public static void main(String[] args) {
Solution s = new Solution();
/**
* 0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0
*/
int[][] board = {
{0, 0, 1, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 1, 0},
{1, 1, 0, 1, 1},
{0, 0, 0, 0, 0},
};
System.out.println(s.hasPath(board, new int[]{0, 4}, new int[]{4, 4}));
/**
* 0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0
*/
int[][] board2 = {
{0, 0, 1, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 1, 0},
{1, 1, 0, 1, 1},
{0, 0, 0, 0, 0},
};
System.out.println(s.hasPath(board2, new int[]{0, 4}, new int[]{3, 2}));
}
public boolean hasPath(int[][] maze, int[] start, int[] dest) {
boolean[][] visited = new boolean[maze.length][maze[0].length];
visited[start[0]][start[1]] = true;
int[][] dirs = {
{0, 1},
{0, -1},
{-1, 0},
{1, 0}
};
Queue<int[]> queue = new LinkedList<>();
queue.add(start);
while (!queue.isEmpty()) {
int[] s = queue.remove();
if (s[0] == dest[0] && s[1] == dest[1]) {
return true;
}
for (int[] dir : dirs) {
int x = s[0] + dir[0];
int y = s[1] + dir[1];
while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
x += dir[0];
y += dir[1];
}
if (!visited[x - dir[0]][y - dir[1]]) {
queue.add(new int[]{x - dir[0], y - dir[1]});
visited[x - dir[0]][y - dir[1]] = true;
}
}
}
return false;
}
}
