# Word Search

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.

Tips:

Code：

``````public class Solution {
public boolean exist(char[][] board, String word) {
int rowLimit = board.length;
int colLimit = board[0].length;
boolean[][] visited = new boolean[rowLimit][colLimit];
for (int i = 0; i < rowLimit; i++) {
for (int j = 0; j < colLimit; j++) {
if (recursive(board, word, 0, i, j, rowLimit, colLimit, visited))
return true;
}
}
return false;
}

public boolean recursive(char[][] board, String word, int index, int row,
int col, int rowLimit, int colLimit, boolean[][] visited) {
if (index == word.length())
return true;
// out of bounds
if (row < 0 || col < 0 || row >= rowLimit || col >= colLimit)
return false;
// at a visited node
if (visited[row][col])
return false;
// the two chars do not match
if (board[row][col] != word.charAt(index))
return false;
// dfs
visited[row][col] = true;
boolean result = recursive(board, word, index + 1, row + 1, col,
rowLimit, colLimit, visited)
|| recursive(board, word, index + 1, row - 1, col, rowLimit,
colLimit, visited)
|| recursive(board, word, index + 1, row, col + 1, rowLimit,
colLimit, visited)
|| recursive(board, word, index + 1, row, col - 1, rowLimit,
colLimit, visited);
visited[row][col] = false;
return result;
}
}
``````