Word Search

WordSearch.java

/**
 * 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.
 */
public class Solution {
    public boolean exist(char[][] board, String word) {
        if(board == null) 
            return false;
        if(word == null || word.length() == 0) 
            return true;
        boolean[][] visited = 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(DFS(board, i, j, word, 0, visited))
                    return true;
            }
        }
        return false;
    }
    private boolean DFS(char[][] board, int i, int j, String word, int len, boolean[][] visited) {
        if(visited[i][j] || board[i][j] != word.charAt(len)) 
            return false;
        if(len == word.length() - 1) 
            return true;
        visited[i][j] = true;
        // walk left;
        if(i != 0 && DFS(board, i - 1, j, word, len + 1, visited))
            return true;
        // walk right;
        if(i != board.length - 1 && DFS(board, i + 1, j, word, len + 1, visited))
            return true;
        // walk up
        if(j != 0 && DFS(board, i, j - 1, word, len + 1, visited))
            return true;
        // walk down
        if(j != board[0].length - 1 && DFS(board, i, j + 1, word, len + 1, visited))
            return true;
        // backtracking
        visited[i][j] = false;
        return false;
    }
}