Word Search

最新推荐文章于 2018-03-15 12:40:13 发布

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本文介绍了一种在二维网格中查找指定单词的算法，利用深度优先搜索（DFS）技术，确保每个单元格只被使用一次。通过实例演示了如何在给定网格中寻找特定单词。

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.

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Have you met this question in a real interview?

思路：很明显的图的遍历，用dfs ，由于每个点只能用一次，所以肯定要一个 visited［］［］作为访问标记。

public class Solution {
public boolean exist(char[][] board, String word) {
    if(word==null || word.length()==0)
        return true;
    if(board==null || board.length==0 || board[0].length==0)
        return false;
    boolean[][] used = new boolean[board.length][board[0].length];
    for(int i=0;i<board.length;i++)
    {
        for(int j=0;j<board[0].length;j++)
        {
            if(search(board,word,0,i,j,used))
                return true;
        }
    }
    return false;
}
private boolean search(char[][] board, String word, int index, int i, int j, boolean[][] used)
{
    if(index == word.length())
        return true;
    if(i<0 || j<0 || i>=board.length || j>=board[0].length || used[i][j] || board[i][j]!=word.charAt(index))
        return false;
    used[i][j] = true;
    boolean res = search(board,word,index+1,i-1,j,used) 
                || search(board,word,index+1,i+1,j,used)
                || search(board,word,index+1,i,j-1,used) 
                || search(board,word,index+1,i,j+1,used);
    used[i][j] = false;
    return res;
}
}