第79題，題目如下:
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 <= 15boardandwordconsists of only lowercase and uppercase English letters.
Follow up: Could you use search pruning to make your solution faster with a larger board?
dfs搜尋到尾的方法很好用。
先確認是不是有找到字，鄰近區域的暴力找法，但是不鄰近就不會去找。一直找上下左右，即可。
“`
public 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, i, j, 0))
return true;
return false;
}
private boolean dfs(char[][] board, String word, int i, int j, int s) {
if (i < 0 || i == board.length || j < 0 || j == board[0].length)
return false;
if (board[i][j] != word.charAt(s) || board[i][j] == ‘*’)
return false;
if (s == word.length() – 1)
return true;
final char cache = board[i][j];
board[i][j] = ‘*’;
final boolean isExist = dfs(board, word, i + 1, j, s + 1) || //
dfs(board, word, i – 1, j, s + 1) || //
dfs(board, word, i, j + 1, s + 1) || //
dfs(board, word, i, j – 1, s + 1);
board[i][j] = cache;
return isExist;
}
“`
