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.
考虑深度优先搜索算法
package 牛客刷题.LeetCode.查找.word_search;
/**
* Created by Administrator on 2018/6/28 0028.
*/
public class Solution {
public static void main(String[] args) {
char[][] board = {{'A','B','C','E'},
{'S','F','C','S'},
{'A','D','E','E'}};
System.out.println(new Solution().exist(board, "ABCCED"));
}
public boolean exist(char[][] board, String word) {
int m = board.length, n = board[0].length;
int[][] visited = new int[m][n]; //0表示还没访问到,1表示已经访问到,不能再用了。
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (solve(board, i, j, word.toCharArray(), visited, 0)) {
return true;
}
}
}
return false;
}
public boolean solve(char[][] board, int i, int j, char[] chars, int[][] visited, int times) {
if (board[i][j] != chars[times]) {
return false;
}
if (times == chars.length -1) {
return true;
}
visited[i][j] = 1;
if (i > 0 && visited[i-1][j] == 0 && solve(board, i-1, j, chars, visited, times +1)) {
return true;
}
if (i < board.length -1 &&visited[i+1][j] == 0 && solve(board, i+1, j, chars, visited, times + 1)) {
return true;
}
if (j > 0 &&visited[i][j-1] == 0 && solve(board, i, j-1, chars, visited, times +1)) {
return true;
}
if (j < board[0].length -1 &&visited[i][j+1] == 0 && solve(board, i, j+1, chars, visited, times + 1)) {
return true;
}
visited[i][j] = 0;
return false;
}
}