leetcode_word_search

发表于

难度：Medium

解题思路：深度优先遍历。如果某个字符串匹配，则对它的上、下、左、右分别进行递归，同时将匹配的字符串辨识成”*”，防止重复，递归好之后，要记得还原。

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class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        bool ret;
        for(int i = 0; i < board.size(); i++)
        {
            for(int j = 0; j < board[0].size(); j++)
            {
                ret = search_char(board,i,j,word,0);
                if(ret ==true)
                    return ret;
            }
        }
        return false;
    }
    bool search_char(vector<vector<char>>& board, int i, int j, string word, int word_index)
    {
        if(i < 0 || i >= board.size())
            return false;
        if(j < 0 || j >= board[0].size())
            return false;
        // cout<<"find: "<<board[i][j]<<endl;
        if(word[word_index] != board[i][j])
        {
            return false;
        }
        else
        {
            char ch = board[i][j];
            board[i][j] = '$';
            bool ret = true;
            if(word_index+1 == word.size())
                return ret;
            ret = ret && search_char(board, i-1,j,word,word_index+1);
            if(ret == true)
                return true;
            ret = true;
            ret = ret && search_char(board, i+1,j,word,word_index+1);
            if(ret == true)
                return true;
            ret = true;
            ret = ret && search_char(board, i,j-1,word,word_index+1);
            if(ret == true)
                return true;
            ret = true;
            ret = ret && search_char(board, i,j+1,word,word_index+1);
            board[i][j] = ch;
            return ret;
        }
    }
};

运行结果：56ms，超过34.01%