Leetcode: Word Search

题目

https://oj.leetcode.com/problems/word-search/

分析

  1. 通过递归来实现的DFS搜索。

  2. 需要注意的一点是, 要用一个visited二维vector来记录已经访问的元素, 以后不能再访问了。

  3. visited形参要声明成引用形式, 每次递归后手动的恢复现场。 如果声明成变量的形式, 会超时。

代码

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

    bool search(vector<vector<char>> &board, int i, int j, string word, int index, vector<vector<bool>> &visited)
    {
        if (index == word.size())
            return true;
        else if (i < 0 || i == board.size() || j < 0 || j == board[0].size())
            return false;
        else if (visited[i][j] == true)
            return false;
        else if (board[i][j] != word[index])
            return false;
        else
        {
            visited[i][j] = true;
            if (search(board, i-1, j, word, index+1, visited)) return true;
            if (search(board, i+1, j, word, index+1, visited)) return true;
            if (search(board, i, j-1, word, index+1, visited)) return true;
            if (search(board, i, j+1, word, index+1, visited)) return true;
            visited[i][j] = false;
        }

        return false;
    }
};

本文章迁移自http://blog.csdn.net/timberwolf_2012/article/details/38826685