Leetcode Word search

最新推荐文章于 2019-05-21 18:57:38 发布

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最新推荐文章于 2019-05-21 18:57:38 发布

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分类专栏： leetcode-dfs leetcode

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本文介绍了一个二维网格中查找特定单词的算法实现。采用深度优先搜索(DFS)策略，递归地从每个可能的起点开始搜索目标单词，确保不重复使用同一字母。通过标记已访问节点避免重复访问。

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,

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

[code]

public class Solution {
    public boolean exist(char[][] board, String word) {
        if(word==null || word.equals(""))return true;
        if(board==null || board.length==0 || board[0].length==0)return false;
        int m=board.length, n=board[0].length;
        char [] cs=word.toCharArray();
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(board[i][j]==cs[0])
                {
                    if(dfs(i,j,board,0,cs))return true;
                }
            }
        }
        return false;
    }

    boolean dfs(int row, int col, char[][]board, int index, char[]cs)
    {
        if(index==cs.length)return true;
        if(row<0 || row>=board.length || col<0 || col>=board[0].length || board[row][col]=='.')return false;

        if(board[row][col]==cs[index])
        {
            board[row][col]='.';
            if(dfs(row+1, col, board, index+1, cs))return true;
            if(dfs(row-1, col, board, index+1, cs))return true;
            if(dfs(row,col+1, board, index+1, cs))return true;
            if(dfs(row, col-1, board, index+1, cs))return true;
            board[row][col]=cs[index];
        }
        return false;
    }
}