2021.10.10 - JZ12.矩阵中的路径

• 1. 题目
• 2. 思路
• • (1) DFS+回溯法
  • (2) DFS+回溯法优化
• 3. 代码

• 遍历矩阵，若元素等于字符串的第一个字符，则从该元素开始进行深度优先搜索。
• 创建一个与原矩阵同等大小的标记矩阵，初始时全部设置为false，遍历过的元素位置设置为true。
• 若相等的字符是字符串的最后一个字符，则表示找到一条符合约束的路径，返回true即可。
• 对当前字符的上下左右四个字符进行递归搜索，符合条件的返回true，否则重新修改标记为false，即回溯。

• 可以直接将原矩阵中遍历过的元素修改为条件之外的字符，回溯时修改为原字符即可。

public class Test {
    public static void main(String[] args) {
        Solution solution = new Solution();
        System.out.println(solution.exist(new char[][]{{'A', 'B', 'C', 'E'}, {'S', 'F', 'E', 'S'}, {'A', 'D', 'E', 'E'}}, "ABCESEEEFS"));
    }
}

class Solution {
    private int row;
    private int col;
    private boolean[][] flag;

    public boolean exist(char[][] board, String word) {
        row = board.length;
        col = board[0].length;
        flag = new boolean[row][col];
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == word.charAt(0)) {
                    if (dfs(board, word, i, j, 0)) {
                        return true;
                    } else {
                        flag[i][j] = false;
                    }
                }
            }
        }
        return false;
    }

    private boolean dfs(char[][] board, String word, int i, int j, int index) {
        flag[i][j] = true;
        if (index == word.length() - 1) {
            return true;
        }
        if (i - 1 >= 0 && !flag[i - 1][j] && board[i - 1][j] == word.charAt(index + 1)) {
            if (dfs(board, word, i - 1, j, index + 1)) {
                return true;
            } else {
                flag[i - 1][j] = false;
            }
        }
        if (i + 1 < row && !flag[i + 1][j] && board[i + 1][j] == word.charAt(index + 1)) {
            if (dfs(board, word, i + 1, j, index + 1)) {
                return true;
            } else {
                flag[i + 1][j] = false;
            }
        }
        if (j - 1 >= 0 && !flag[i][j - 1] && board[i][j - 1] == word.charAt(index + 1)) {
            if (dfs(board, word, i, j - 1, index + 1)) {
                return true;
            } else {
                flag[i][j - 1] = false;
            }
        }
        if (j + 1 < col && !flag[i][j + 1] && board[i][j + 1] == word.charAt(index + 1)) {
            if (dfs(board, word, i, j + 1, index + 1)) {
                return true;
            } else {
                flag[i][j + 1] = false;
            }
        }
        return false;
    }
}

class Solution1 {
    private int row;
    private int col;

    public boolean exist(char[][] board, String word) {
        row = board.length;
        col = board[0].length;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == word.charAt(0)) {
                    if (dfs(board, word, i, j, 0)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }

    private boolean dfs(char[][] board, String word, int i, int j, int index) {
        if (i < 0 || i >= row || j < 0 || j >= col || board[i][j] != word.charAt(index)) {
            return false;
        }
        if (index == word.length() - 1) {
            return true;
        }
        board[i][j] = '';
        boolean res = dfs(board, word, i - 1, j, index + 1) || dfs(board, word, i + 1, j, index + 1) || dfs(board, word, i, j - 1, index + 1) || dfs(board, word, i, j + 1, index + 1);
        board[i][j] = word.charAt(index);
        return res;
    }
}