难度meidum
给定一个二维网格和一个单词,找出该单词是否存在于网格中。
单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
示例:
board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
给定 word = "ABCCED", 返回 true
给定 word = "SEE", 返回 true
给定 word = "ABCB", 返回 false
提示:
board 和 word 中只包含大写和小写英文字母。
1 <= board.length <= 200
1 <= board[i].length <= 200
1 <= word.length <= 10^3
解题思路:深度优先搜索
代码 t50 s90 java
class Solution { public boolean exist(char[][] board, String word) { int m = board.length, n = board[0].length; for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(dfs(board, word, i, j, 0)) return true; } } return false; } public boolean dfs(char[][] board, String word, int i, int j, int idx){ if(idx==word.length()) return true; if(i<0 || j<0 || i>=board.length || j>=board[0].length || board[i][j]!=word.charAt(idx)) return false; char t = board[i][j]; board[i][j] = '#'; boolean res = dfs(board, word, i+1, j, idx+1) || dfs(board, word, i, j+1, idx+1) || dfs(board, word, i-1, j, idx+1) || dfs(board, word, i, j-1, idx+1); board[i][j] = t; return res; } }
参考资料:
http://www.cnblogs.com/grandyang/p/4332313.html