Leetcode 79. Word Search

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本文介绍了一种在二维字符网格中查找指定单词的算法实现。该算法通过深度优先搜索(DFS)递归地检查每个可能的路径，确保同一字母单元格不被重复使用。文章提供了一个Java实现示例，并展示了如何通过递归函数判断单词是否存在。

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Question:

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 =

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

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

Solution:

package algorithm;

/**
 * @author

 * @version V 1.0 2016年4月11日
 * @since jdk1.6
 */

//	79. Word Search

public class WordSearch {
	public boolean exist(char[][] board, String word) {
		boolean flag = false;

		int index = 0;

		for (int row = 0; row < board.length; row++)
			for (int rank = 0; rank < board[row].length; rank++) {
				if (board[row][rank] == word.charAt(index)) {
//					int[][] idt = new int[board.length][board[0].length];
					flag = calc(row, rank, index, board, word);
					if (flag)
						return true;
				}
			}
		return flag;
	}

	public boolean calc(int row, int rank, int index, char[][] board, String word) {
		char tem = 0 ;
		if (row < 0 || rank < 0 || row >= board.length || rank >= board[row].length || index >= word.length()
				|| board[row][rank] != word.charAt(index)) {
			return false;
		} 
		else if (index == word.length() - 1) {
			return true;
		} else {
			tem = board[row][rank];
			board[row][rank] = '-';
			boolean flag = (calc(row - 1, rank, index + 1, board, word)
					|| calc(row + 1, rank, index + 1, board, word) || calc(row, rank - 1, index + 1, board, word) || calc(
					row, rank + 1, index + 1, board, word));
			if (!flag)
				board[row][rank] = tem;
			return flag;
		}
	}

	public static void main(String[] args) {
		char[][] board = { { 'a', 'a' }};
		String word = "aaa";
/*		char[][] board = { { 'a', 'b' }, { 'c', 'd' } };
		String word = "cdba";
		
*/		/*		char [][] board = {{'A','B','C','E'},{'S','F','C','S'},{'A','D','E','E'}};
				String word = "ABCCED";
		*/System.out.println(new WordSearch().exist(board, word));
	}
}