剑指 Offer 12. 矩阵中的路径

题目链接

• 题目类似于迷宫的搜索。
• 需要注意的是，需要首先判断起始搜索的位置，可能有多个起点，都需要一一尝试。
• 每轮迭代的时候记得将是否遍历标记数组还原为未遍历的状态。

package com.walegarrett.offer;

/**
 * @Author WaleGarrett
 * @Date 2020/12/9 9:09
 */

import java.util.Arrays;

/**
 * [["a","b","c","e"],
 * ["s","f","c","s"],
 * ["a","d","e","e"]]
 * 矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。
 * 1 <= board.length <= 200
 * 1 <= board[i].length <= 200
 */
public class Offer_12 {
    private String word;
    private boolean[][] travel;
    private final int[][] direction = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public boolean exist(char[][] board, String word) {
        if(board == null)
            return false;
        if(word.length() == 0)
            return true;
        this.word = word;
        travel = new boolean[board.length][board[0].length];
        for(int i =0; i< board.length; i++){
            for(int j =0; j< board[0].length; j++){
                travel[i][j] = false;
            }
        }
        for(int i =0; i< board.length; i++){
            int len = board[0].length;
            for(int j =0; j< len; j++){
                if(board[i][j] != word.charAt(0))
                    continue;
                travel[i][j] = true;
                if(findPath(board, 0, i, j)){
                    return true;
                }
                travel[i][j] = false;
            }
        }
        return false;
    }
    public boolean findPath(char[][] board, int pos, int x, int y){
        if(pos >= word.length() - 1){
            if(board[x][y] == word.charAt(pos))
                return true;
            return false;
        }
        //遍历四个方向
        for(int i = 0; i< 4; i++){
            int dx = x + direction[i][0];
            int dy = y+ direction[i][1];
            if(dx>=0 && dy>=0 && dx<board.length && dy< board[0].length && !travel[dx][dy]){
                if(board[dx][dy] == word.charAt(pos + 1)){
                    travel[dx][dy] = true;
                    if(findPath(board, pos+1, dx, dy)){
                        travel[dx][dy] = false;
                        return true;
                    }else travel[dx][dy] = false;
                }
            }
        }
        return false;
    }
}

剑指 Offer 12. 矩阵中的路径 + 递归 + 深搜 + 字符串问题

原文：https://www.cnblogs.com/GarrettWale/p/14106968.html