79. Word Search
Medium
Given an m x n grid of characters board and a string word, return true if word exists in the grid.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
Example 1:
-
Input:
board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" -
Output:
true
Example 2:
-
Input:
board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" -
Output:
true
Example 3:
-
Input:
board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" -
Output:
false
Constraints:
m == board.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15-
boardandwordconsists of only lowercase and uppercase English letters.
Follow-up: Could you use search pruning to make your solution faster with a larger board?
Solution:
class Solution {
/**
* @param String[][] $board
* @param String $word
* @return Boolean
*/
function exist($board, $word) {
for ($i = 0; $i < count($board); ++$i)
for ($j = 0; $j < count($board[0]); ++$j)
if ($this->dfs($board, $word, $i, $j, 0))
return true;
return false;
}
private function dfs($board, $word, $i, $j, $s)
{
if ($i < 0 || $i == count($board) || $j < 0 || $j == count($board[0]))
return false;
if ($board[$i][$j] != $word[$s] || $board[$i][$j] == '*')
return false;
if ($s == strlen($word) - 1)
return true;
$cache = $board[$i][$j];
$board[$i][$j] = '*';
$isExist = $this->dfs($board, $word, $i + 1, $j, $s + 1) ||
$this->dfs($board, $word, $i - 1, $j, $s + 1) ||
$this->dfs($board, $word, $i, $j + 1, $s + 1) ||
$this->dfs($board, $word, $i, $j - 1, $s + 1);
$board[$i][$j] = $cache;
return $isExist;
}
}
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