给定一个二维面板和一个单词,找出该单词是否存在于网格中。
这个词可由顺序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。
例如,
给定 二维面板 =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]
单词= "ABCCED",-> 返回 true,
单词 = "SEE", -> 返回 true,
单词 = "ABCB", -> 返回 false。
详见:https://leetcode.com/problems/word-search/description/

Java实现:

class Solution {
public boolean exist(char[][] board, String word) {
int m = board.length;
int n = board[0].length;
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (dfs(board, word, 0, i, j, visited)) {
return true;
}
}
}
return false;
} public boolean dfs(char[][] board, String word, int index, int rowindex, int colindex, boolean[][] visited) {
if (index == word.length()){
return true;
}
if (rowindex < 0 || colindex < 0 || rowindex >=board.length || colindex >= board[0].length){
return false;
}
if (visited[rowindex][colindex]){
return false;
}
if (board[rowindex][colindex] != word.charAt(index)){
return false;
}
visited[rowindex][colindex] = true;
boolean res =
dfs(board, word, index + 1, rowindex - 1, colindex, visited)
|| dfs(board, word, index + 1, rowindex + 1, colindex, visited)
|| dfs(board, word, index + 1, rowindex, colindex + 1, visited)
|| dfs(board, word, index + 1, rowindex, colindex - 1, visited);
visited[rowindex][colindex] = false;
return res;
}
}

参考:https://www.cnblogs.com/springfor/p/3883942.html

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