【LeetCode】79. 单词搜索

题目

给定一个二维网格和一个单词，找出该单词是否存在于网格中。
单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.

思路

同 【剑指Offer】面试题12. 矩阵中的路径
以每个位置开头检查是否存在路径。

代码

时间复杂度：O(n * m)
空间复杂度：O(1)

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

    bool dfs(vector<vector<char>> &board, int i, int j, int len, string word) {     
        int row = board.size(), col = board[0].size();
        if (i < 0 || i >= row || j < 0 || j >= col || word[len] != board[i][j]) return false;      
        if (len == word.size() - 1) return true;
        ++len;
        char ch = board[i][j];
        board[i][j] = '#';
        bool ret = dfs(board, i - 1, j, len, word) ||
                   dfs(board, i + 1, j, len, word) ||
                   dfs(board, i, j - 1, len, word) ||
                   dfs(board, i, j + 1, len, word);        
        board[i][j] = ch; //回溯
        return ret;
    }
};

另一种写法

使用访问数组表示每个位置是否访问。
时间复杂度：O(n * m)
空间复杂度：O(n * m)

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

    bool dfs(vector<vector<char>> &board, int i, int j, int len, string word, vector<vector<bool>> &visited) {     
        int row = board.size(), col = board[0].size();
        if (i < 0 || i >= row || j < 0 || j >= col || visited[i][j] || word[len] != board[i][j]) return false;      
        if (len == word.size() - 1) return true;
        ++len;
        visited[i][j] = true;
        bool ret = dfs(board, i - 1, j, len, word, visited) ||
                   dfs(board, i + 1, j, len, word, visited) ||
                   dfs(board, i, j - 1, len, word, visited) ||
                   dfs(board, i, j + 1, len, word, visited);        
        visited[i][j] = false; //回溯
        return ret;
    }
};