题目：
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.
思路：
递归；及时return true，减少计算量
package recursion; public class WordSearch { public boolean exist(char[][] board, String word) { int m; int n; int wordLen; if (board == null || word == null || (m = board.length) == 0 || (n = board[0].length) == 0 || (wordLen = word.length()) == 0 || wordLen > m * n) return false; boolean[][] visited = new boolean[m][n]; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (dfs(board, i, j, visited, word, 0, m, n, wordLen)) return true; } } return false; } private boolean dfs(char[][] board, int x, int y, boolean[][] visited, String word, int pos, int m, int n, int wordLen) { if (visited[x][y] || board[x][y] != word.charAt(pos)) return false; if (pos == wordLen - 1) return true; visited[x][y] = true; if (x - 1 >= 0 && dfs(board, x - 1, y, visited, word, pos + 1, m, n, wordLen)) return true; if (x + 1 < m && dfs(board, x + 1, y, visited, word, pos + 1, m, n, wordLen)) return true; if (y - 1 >= 0 && dfs(board, x, y - 1, visited, word, pos + 1, m, n, wordLen)) return true; if (y + 1 < n && dfs(board, x, y + 1, visited, word, pos + 1, m, n, wordLen)) return true; visited[x][y] = false; return false; } public static void main(String[] args) { // TODO Auto-generated method stub char[][] board = { {'A','B','C','E'}, {'S','F','C','S'}, {'A','D','E','E'} }; WordSearch w = new WordSearch(); System.out.println(w.exist(board, "ABCCED")); System.out.println(w.exist(board, "SEE")); System.out.println(w.exist(board, "ABCB")); } }