Sword refers to Offer 12. The path in the matrix-backtracking DFS

Please design a function to determine whether there is a path that contains all the characters of a string in a matrix. The path can start from any grid in the matrix, and each step can move one grid to the left, right, up, and down in the matrix. If a path passes through a grid of the matrix, then the path cannot enter the grid again. For example, the following 3×4 matrix contains a path for the character string "bfce" (the letters in the path are marked in bold).

[[“a”,“b”,“c”,“e”],
[“s”,“f”,“c”,“s”],
[“a”,“d”,“e”,“e”]]

But the matrix does not contain the path of the string "abfb", because after the first character b of the string occupies the second grid in the first row of the matrix, the path cannot enter this grid again.

Example 1:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D", "E","E"]], word = "ABCCED"
Output: true
Example 2:

Input: board = [["a","b"],["c","d"]], word = "abcd"
Output: false

Source: LeetCode
Link: Click to jump to https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof

//回溯法，又称深度优先搜索（DFS）
class Solution {
    
    
    //函数返回是否存在该路径
    public boolean exist(char[][] board, String word) {
    
    
        char[] ch = word.toCharArray();
        //判断位置是否被走过，默认false：没走过
        boolean[][] isused = new boolean[board.length][board[0].length];
        //循环遍历二维数组每个位置
        for(int i = 0; i < board.length; i++) {
    
    
            for(int j = 0; j < board[0].length; j++) {
    
    
                //然后每个位置判断是否可以达成word路径
                if(dfs(board, ch, isused, i, j, 0)) return true;
            }
        }
        return false;
    }

    //递归每个位置，回溯法：先朝一个方向搜到底，再回溯至上个节点，沿另一个方向搜索，以此类推
    public boolean dfs(char[][] board, char[] ch, boolean[][] isused, int i, int j, int index) {
    
    
        //不符合word路径的条件
        //1.该位置越界
        //2.board(i,j)位置不符合word(index)位置
        //3.该位置已经走过
        if(i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != ch[index] || isused[i][j]) {
    
    
            return false;
        }
        //符合word路径的条件:
        if(index == ch.length-1) {
    
    
            return true;
        }
        //已经走过这个位置了
        isused[i][j] = true;
        //step先向上一直搜索，不行了再回溯上一个节点，再向下。。。再向左。。。再向右。。。
        boolean step = false;
        step = dfs(board, ch, isused, i-1, j, index+1) ||
                dfs(board, ch, isused, i+1, j, index+1) ||
                dfs(board, ch, isused, i, j-1, index+1) ||
                dfs(board, ch, isused, i, j+1, index+1);
        //因为走不下去，退回到上一位置了，所以该位置重置为false：没走过
        isused[i][j] = false;
        return step;
    }
}