题目：（Backtrancing）
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.
题解：
比较有意思的backtracing.
public class Solution{ public static boolean exist(char[][] board, String word) { for(int i=0; i<board.length; i++) { for(int j=0; j<board[0].length; j++) { if(dfs(board, word, 0, i, j)) return true; } } return false; } public static boolean dfs(char[][] board, String word, int index, int x, int y){ if(index == word.length()-1 && word.charAt(index)==board[x][y]) return true; if(word.charAt(index) != board[x][y]) return false; char tmp = board[x][y]; board[x][y] = ‘.‘; boolean b1 = false, b2 = false, b3 = false, b4 = false; if(x-1>=0 && board[x-1][y] != ‘.‘) b1 = dfs(board, word, index+1, x-1, y); if(!b1 && y-1>=0 && board[x][y-1] != ‘.‘) b2 = dfs(board, word, index+1, x, y-1); if(!b1 && !b2 && x+1<board.length && board[x+1][y] != ‘.‘) b3 = dfs(board, word, index+1, x+1, y); if(!b1 && !b2 && !b3 && y+1<board[0].length && board[x][y+1] != ‘.‘) b4 = dfs(board, word, index+1, x, y+1); board[x][y] = tmp; return b1 || b2 || b3 || b4; } }
原文：http://www.cnblogs.com/fengmangZoo/p/4212071.html