剑指 offer 12. 矩阵中的路径

剑指 Offer 12. 矩阵中的路径

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但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。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["a","b"],["c","d"]], word = "abcd"
输出：false

提示：

• 1 <= board.length <= 200
• 1 <= board[i].length <= 200

注意：本题与主站 79 题相同：https://leetcode-cn.com/problems/word-search/

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution 1

递归查找
用 ‘.’ 标记已经查找过的

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

    private boolean recursivelyFind(char[][] board, int i, int j, String word, int pos) {
        if (pos == word.length())
            return true;
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length)
            return false;
        char c = word.charAt(pos);
        boolean found = false;
        if (board[i][j] == c) {
            board[i][j] = '.';
            found = recursivelyFind(board, i - 1, j, word, pos + 1)
                 || recursivelyFind(board, i, j - 1, word, pos + 1)
                 || recursivelyFind(board, i + 1, j, word, pos + 1)
                 || recursivelyFind(board, i, j + 1, word, pos + 1);
            board[i][j] = c;
        }
        return found;
    }
}

Solution 2

cpp

class Solution
{
public:
    bool exist(vector<vector<char>> &board, string word)
    {
        if (word.length() == 0)
        {
            return true;
        }
        rows = board.size();
        if (rows == 0)
        {
            return false;
        }
        columns = board[0].size();
        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < columns; j++)
            {
                if (find(board, i, j, word, 0))
                {
                    return true;
                }
            }
        }
        return false;
    }

    bool find(vector<vector<char>> &board, int i, int j, string word, int k)
    {
        if (i < 0 || i >= rows || j < 0 || j >= columns || board[i][j] != word[k])
        {
            return false;
        }
        if (k == word.length() - 1)
        {
            return true;
        }
        board[i][j] = '\0';
        bool found = find(board, i + 1, j, word, k + 1) || find(board, i - 1, j, word, k + 1) || find(board, i, j + 1, word, k + 1) || find(board, i, j - 1, word, k + 1);
        board[i][j] = word[k];
        return found;
    }

protected:
    int rows = 0, columns = 0;
};