Word Search
问题：
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.
思路：
dfs + 回溯
我的代码：
public class Solution { public boolean exist(char[][] board, String word) { if(board == null || board.length == 0 || board[0].length == 0 || word == null || word.length() == 0) return false; int row = board.length; int col = board[0].length; boolean[][] isVisited = new boolean[row][col]; for(int i = 0; i < row; i++) { for(int j = 0; j < col; j++) { if(board[i][j] == word.charAt(0)) { isVisited[i][j] = true; if(helper(board, word.substring(1), i, j, isVisited, row, col)) return true; isVisited[i][j] = false; } } } return false; } public boolean helper(char[][] board, String word, int i, int j, boolean[][] isVisited, int row, int col) { if(word == null || word.length() == 0) return true; char c = word.charAt(0); //left if(j > 0 && isVisited[i][j-1] == false && board[i][j-1] == c) { isVisited[i][j-1] = true; if(helper(board, word.substring(1), i, j-1, isVisited, row, col)) return true; isVisited[i][j-1] = false; } //right if(j < col - 1 && isVisited[i][j+1] == false && board[i][j+1] == c) { isVisited[i][j+1] = true; if(helper(board, word.substring(1), i, j+1, isVisited, row, col)) return true; isVisited[i][j+1] = false; } //up if(i > 0 && isVisited[i-1][j] == false && board[i-1][j] == c) { isVisited[i-1][j] = true; if(helper(board, word.substring(1), i-1, j, isVisited, row, col)) return true; isVisited[i-1][j] = false; } //down if(i < row-1 && isVisited[i+1][j] == false && board[i+1][j] == c) { isVisited[i+1][j] = true; if(helper(board, word.substring(1), i+1, j, isVisited, row, col)) return true; isVisited[i+1][j] = false; } return false; } }
原文：http://www.cnblogs.com/sunshisonghit/p/4357290.html
