小红书：笔试题（棋盘最短路径，笔记本草稿栈，迷宫游戏）
1. 棋盘最短路径问题
题目描述：
题目描述： 假设以一个n*m的矩阵作为棋盘，每个棋位对应一个二维坐标 (x, y)。你有一颗棋子位于左上起点(0, 0)，现在需要将其移动到右下底角 (n-1, m-1)，棋子可以向相邻的上下左右位置移动，每个坐标最多只能经过一次。棋盘中散布着若干障碍，障碍物不能跨越，只能绕行，问是否存在到达右下底角的路线？若存在路线，输出所需的最少移动次数；若不存在，输出0。 Input 第一行三个正整数n，m和k，代表棋盘大小与障碍物个数 1< n、m < 100, k < n*m 第二行至第k+1行，每行为两个整数x和y，代表k个障碍物的坐标。 输入 输入三个正整数n，m和k，代表棋盘大小与障碍物个数 1< n、m < 100, k < n*m 第二行至第k+1行，每行为两个整数x和y，代表k个障碍物的坐标。 输出 输出从起点到终点的最短路径的长度，如果不存在，即输出0
代码：
import java.util.Arrays; import java.util.Scanner; public class Main { public static int min = Integer.MAX_VALUE; public static boolean find = false; public static void main( String[] args ) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt();//障碍 boolean[][] go = new boolean[n][m];//棋盘 boolean[][] visit = new boolean[n][m];//走过的痕迹 //将所有的位置设置为true for(int i=0;i<n;i++) { Arrays.fill(go[i], true); } //有棋子障碍处设置为false for(int i=0;i<k;i++) { int a = sc.nextInt(); int b = sc.nextInt(); go[a][b] = false; } findPath(0, 0, 0, visit, go, n - 1, m - 1); //如果找到了输出结果，没路输出0 if(find) { System.out.println(min); }else { System.out.println(0); } } private static void findPath( int row , int col , int s, boolean[][] visit, boolean[][] go, int targetRow, int targetCol ) { if(row == targetRow && col == targetCol) {//到达目标位置 find = true; if(s < min) { min = s; } return; } if(col < targetCol && !visit[row][col+1] && go[row][col+1]) { // 右 visit[row][col+1] = true; findPath(row, col + 1, s + 1, visit, go, targetRow, targetCol); visit[row][col+1] = false; } if(row < targetRow && !visit[row+1][col] && go[row+1][col]) { // 下 visit[row+1][col] = true; findPath(row + 1, col, s + 1, visit, go, targetRow, targetCol); visit[row+1][col] = false; } if(col > 0 && !visit[row][col-1] && go[row][col-1]) { // 左 visit[row][col-1] = true; findPath(row, col - 1, s + 1, visit, go, targetRow, targetCol); visit[row][col-1] = false; } if(row > 0 && !visit[row-1][col] && go[row-1][col]) { // 上 visit[row-1][col] = true; findPath(row - 1, col, s + 1, visit, go, targetRow, targetCol); visit[row-1][col] = false; } } }
2. 笔记草稿
代码1,利用栈
import java.util.Scanner; import java.util.Stack; public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); String str = sc.nextLine(); String results = getRealContent(str); System.out.println(results); } public static String getRealContent(String str) { StringBuilder sb = new StringBuilder(""); char flag = '('; Stack<Character> stack = new Stack<Character>(); char[] chars = str.toCharArray(); //遍历每个数组将其放入栈中 for (int i = 0; i < chars.length; i++) { if (chars[i] =='(') {//入栈 stack.push(chars[i]); }else if (chars[i] ==')') {//弹出和这个)对应的(以及他们中间的内容 while(!stack.isEmpty() && stack.peek()!= flag){ stack.pop(); } stack.pop(); }else if (chars[i] =='<') {//删除前面的一个内容 if(!stack.empty() && stack.peek()!=flag){ stack.pop(); } }else{ stack.push(chars[i]); } } while(!stack.empty()){ sb.append(stack.pop()); } return sb.reverse().toString(); } }
代码2，括号加减
import java.util.Scanner; public class Main { public static void main( String[] args ) { Scanner sc = new Scanner(System.in); String s = sc.next(); String res = ""; int count = 0, len = s.length(); for(int i=0;i<len;i++) { char c = s.charAt(i); if(c == '(') { count ++; }else if(c == ')') { count --; }else if(c == '<') { if(res.length() > 0 && count == 0 && res.charAt(res.length() - 1) != ')') { res = res.substring(0, res.length() - 1); } }else if(count == 0) { res += c; } } System.out.println(res); } }
3. 迷宫游戏
import java.util.*; /** * 5 * .#... * ..#S. * .E### * ..... * ..... */ public class B { static int[][] d = { {-1, 0}, {0, 1}, {1, 0}, {0, -1} }; static int startX = 0; static int startY = 0; static int endX = 0; static int endY = 0; static Set<Integer> stepList = new HashSet<>(); public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); in.nextLine(); char[][] board = new char[n][n]; for (int i = 0; i < n; i++) { String line = in.nextLine(); for (int j = 0; j < n; j++) { board[i][j] = line.charAt(j); if (board[i][j] == 'S') { startX = i; startY = j; } else if (board[i][j] == 'E') { endX = i; endY = j; } } } shortestLength(board, n); if (stepList.size() == 0) { System.out.println(-1); return; } int minStep = Integer.MAX_VALUE; for (Integer integer : stepList) { if (integer < minStep) { minStep = integer; } } System.out.println(minStep); } private static void shortestLength(char[][] board, int n) { boolean[][] visited = new boolean[n][n]; visited[startX][startY] = true; shortestLength(board, n, startX, startY, 0, visited); } private static void shortestLength(char[][] board, int n, int startX, int startY, int step, boolean[][] visited) { if (startX == endX && startY == endY) { stepList.add(step); return; } for (int i = 0; i < 4; i++) { int newX = startX + d[i][0]; int newY = startY + d[i][1]; if (newX < 0) { newX = n - 1; } else if (newX==n) { newX = 0; } if (newY <0) { newY = n-1; }else if (newY==n){ newY = 0; } if (board[newX][newY] !='#' && !visited[newX][newY]) { visited[newX][newY] = true; shortestLength(board, n, newX, newY, step + 1, visited); visited[newX][newY] = false; } } } }
