LeetCode 79: Word Search

最新推荐文章于 2020-06-15 16:03:04 发布

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本文介绍了一种算法，用于在给定的二维网格中查找特定单词的路径。算法通过深度优先搜索（DFS）来遍历相邻的单元格，确保不重复使用相同的字母单元格。

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Difficulty: 3

Frequency: 4

Problem:

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.

For example,
Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

Solution:

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (word.size()==0)
            return true;
        
        if (word.size()==1)
        {
            for (int i = 0; i<board.size(); i++)
            {
                for (int j = 0; j<board[i].size(); j++)
                {
                    if (board[i][j]==word[0])
                        return true;
                }
            }
            return false;
        }
        
        vector<vector<char> > visited(board.size());
        for (int i = 0; i<visited.size(); i++)
        {
            visited[i].resize(board[i].size(), 0);
        }
        for (int i = 0; i<board.size(); i++)
        {
            for (int j = 0; j<board[i].size(); j++)
            {
                if (board[i][j]==word[0]&&DFS(board, visited, i, j, word, 1))
                    return true;
            }
        }
        return false;
    }
    bool DFS (vector<vector<char> > & board, vector<vector<char> > & visited, int row, int col, const string & word, int i_word)
    {
        visited[row][col] = 1;
        if (row!=0&&!visited[row-1][col])
        {
            if (board[row-1][col]==word[i_word])
            {
                if (i_word==word.size()-1)
                    return true;
                else if (DFS(board, visited, row-1, col, word, i_word+1))
                    return true;
            }
                
        }
        if (row<visited.size()-1&&!visited[row+1][col])
        {
            if (board[row+1][col]==word[i_word])
            {
                if (i_word==word.size()-1)
                    return true;
                else if (DFS(board, visited, row+1, col, word, i_word+1))
                    return true;
            }
        }
        if (col!=0&&!visited[row][col-1])
        {
            if (board[row][col-1]==word[i_word])
            {
                if (i_word==word.size()-1)
                    return true;
                else if (DFS(board, visited, row, col-1, word, i_word+1))
                    return true;
            }
        }
        if (col<visited[row].size()-1&&!visited[row][col+1])
        {
            if (board[row][col+1]==word[i_word])
            {
                if (i_word==word.size()-1)
                    return true;
                else if (DFS(board, visited, row, col+1, word, i_word+1))
                    return true;
            }
        }
        
        visited[row][col] = 0;
        return false;
    }
};

Notes:

DFS.