题目描述：
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.
代码实现：
class Solution {
public static boolean exist(char[][] board, String word) {
if (board.length == 0) {
return false;
}
//网格的行数
int row = board.length;
//网格的列数
int col = board[0].length;
//表示网格是否已经被访问过的状态
boolean[][] visited = new boolean[row][col];
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (board[i][j] == word.charAt(0) && backTrace(i, j, 0, board, word, visited)) {
return true;
}
}
}
return false;
}
//DFS+回溯
private static boolean backTrace(int i, int j, int index, char[][] board, String word, boolean[][] visited) {
if (index == word.length()) {
return true;
}
//二维平面按照上->右->下->左的顺序搜索
if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != word.charAt(index) || visited[i][j]) {
return false;
}
//标记当前网格已经被访问
visited[i][j] = true;
if (backTrace(i - 1, j, index + 1, board, word, visited)
|| backTrace(i, j + 1, index + 1, board, word, visited)
|| backTrace(i + 1, j, index + 1, board, word, visited)
|| backTrace(i, j - 1, index + 1, board, word, visited)) {
return true;
}
//回溯
visited[i][j] = false;
return false;
}
}
原文：https://www.cnblogs.com/ysw-go/p/11818677.html