【CodeWars】Path Finder #2: shortest path

题意

题目链接：https://www.codewars.com/kata/path-finder-number-2-shortest-path

Task
You are at position [0, 0] in maze NxN and you can only move in one of the four cardinal directions (i.e. North, East, South, West). Return the minimal number of steps to exit position [N-1, N-1] if it is possible to reach the exit from the starting position. Otherwise, return false in JavaScript/Python and -1 in C++/C#/Java.Empty positions are marked .. Walls are marked W. Start and exit positions are guaranteed to be empty in all test cases.

题目意思很清楚，就是求解迷宫的最短路径长度，迷宫格式如下，点表示路，W表示墙，入口是(0,0)位置，出口是最后一行的最后一个点：

".W.\n“”
".W.\n""..."

代码

迷宫有很多求解方法，我采用的是BFS，代码中搜索路径经过的点，其中(x,y)是左边，steps表示到达当前位置经过的步数，当搜索到迷宫出口时放回对应的steps就可以了，需要注意的是BFS使用了队列，需要标记某个位置是否将会被访问，也就是是否已经加入到队列中（这里使用steps作为标记，当steps为0时表示未加入队列，这样可以省去一个变量），需要在坐标入队时将对应的steps标记为上一步steps+1，而不是在坐标出队的时候标记访问，否则会重复搜索很多的位置，导致时间复杂度爆炸

import java.util.LinkedList;
import java.util.Queue;public class Finder {public static int pathFinder(String maze) {maze = maze.replaceAll("\n","");StringBuilder stringBuilder = new StringBuilder(maze);int n = (int) Math.sqrt(maze.length());stringBuilder.replace(0,1,"W");Queue<Position> positionQueue = new LinkedList<Position>() {{offer(new Position(0,0,0));}};int[][] dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};while (!positionQueue.isEmpty()) {Position position = positionQueue.poll();int i = position.x;int j = position.y;if (i == n - 1 && j == n - 1) return position.steps;for (int k = 0; k < dir.length; k++) {int posX = i + dir[k][0];int posY = j + dir[k][1];if (posX < 0 || posY < 0 || posX >= n || posY >= n || stringBuilder.charAt(posX * n + posY) == 'W')continue;positionQueue.offer(new Position(posX,posY,position.steps + 1));stringBuilder.replace(posX * n + posY,posX * n + posY + 1,"W");}}return -1;}static class Position{int x;int y;int steps;public Position(int x, int y, int steps) {this.x = x;this.y = y;this.steps = steps;}}
}

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【CodeWars】Path Finder #2: shortest path

题意

代码

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