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.dfs + backtracing
首先遍历矩阵中的每个元素，是否等于word的第一个字符，如果等于进行dfs。在ｄｆｓ中，如果当前元素周围的元素未被遍历过、并且等于word里的下一个字符，就再对这个元素进行dfs（ｂａｃｋｔｒａｃｉng）
这是２１２的先头题。
public class Solution {
public boolean exist(char[][] board, String word) {
if (word == null || word.length() == 0) {
return false;
}
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 (board[i][j] == word.charAt(0)) {
if (helper(visited, board, i, j, word, 1)) {
return true;
}
}
}
}
return false;
}
public boolean helper(boolean[][] visited, char[][] board, int i, int j, String word, int index) {
if (index == word.length()) {
return true;
}
visited[i][j] = true;
boolean re = false;
if (!re && i > 0 && board[i - 1][j] == word.charAt(index) && !visited[i - 1][j]) {
re = helper(visited, board, i - 1, j, word, index + 1);
}
if (!re && i < board.length - 1 && board[i + 1][j] == word.charAt(index) && !visited[i + 1][j]) {
re = helper(visited, board, i + 1, j, word, index + 1);
}
if (!re && j > 0 && board[i][j - 1] == word.charAt(index) && !visited[i][j - 1]) {
re = helper(visited, board, i, j - 1, word, index + 1);
}
if (!re && j < board[0].length - 1 && board[i][j + 1] == word.charAt(index) && !visited[i][j + 1]) {
re = helper(visited, board, i, j + 1, word, index + 1);
}
visited[i][j] = false;
return re;
}
}