79 Word Search

最新推荐文章于 2023-12-29 13:37:00 发布

weixin_39145266

最新推荐文章于 2023-12-29 13:37:00 发布

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分类专栏： # Array LeetCode # Medium # Solved # Backtracking

本文链接：https://blog.youkuaiyun.com/weixin_39145266/article/details/89184293

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本文介绍了一种在二维网格中查找特定单词的算法实现。通过深度优先搜索(DFS)策略，算法检查网格中的每个单元格，寻找与目标单词匹配的路径。文章详细解释了递归搜索过程，并提供了C++代码示例，展示了如何避免重复使用同一字母单元格。

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Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example:

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

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

2 尝试解

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(word.size() == 0) return true;
        if(board.size()==0|| board[0].size()==0 ) return false;
        for(int i = 0; i < board.size();i++){
            for(int j = 0; j < board[0].size();j++){
                if(board[i][j] == word[0]){
                    board[i][j] = '0';
                    if(find(board,word.substr(1),i,j)) return true;
                    board[i][j] = word[0];
                }
            }
        }
        return false;

    }
    bool find(vector<vector<char>>& board, string word,int row,int column){
        if(word.size()==0) return true;
        if(row>0 && board[row-1][column] == word[0]){
            board[row-1][column] = '0';
            if(find(board,word.substr(1),row-1,column)) return true;
            board[row-1][column] = word[0];
        }
        if(column>0 && board[row][column-1] == word[0]){
            board[row][column-1] = '0';
            if(find(board,word.substr(1),row,column-1)) return true;
            board[row][column-1] = word[0];
        }
        if(row < board.size()-1 && board[row+1][column] == word[0]){
            board[row+1][column] = '0';
            if(find(board,word.substr(1),row+1,column)) return true;
            board[row+1][column] = word[0];
        }
        if((column < board[0].size()-1) && board[row][column+1] == word[0]){
            board[row][column+1] = '0';
            if(find(board,word.substr(1),row,column+1)) return true;
            board[row][column+1] = word[0];
        }
        return false;
    }
};

3 标准解

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if (word.empty()) return true;
        if (board.empty() || board[0].empty()) return false;
        
        m = board.size();
        n = board[0].size();
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                // traverse the board to find the first char
                if (dfsSearch(board, word, 0, i, j)) return true;
        return false;
    }
private:
    int m;
    int n;
    bool dfsSearch(vector<vector<char>>& board, string& word, int k, int i, int j) {
        if (i < 0 || i >= m || j < 0 || j >= n || word[k] != board[i][j]) return false;
        if (k == word.length() - 1) return true;  // found the last char

        char cur = board[i][j];
        board[i][j] = '*';  // used
        bool search_next = dfsSearch(board, word, k+1, i-1 ,j) 
                        || dfsSearch(board, word, k+1, i+1, j) 
                        || dfsSearch(board, word, k+1, i, j-1)
                        || dfsSearch(board, word, k+1, i, j+1);
        board[i][j] = cur;  // reset
        return search_next;
    }
};