#yyds干货盘点# LeetCode面试题：单词搜索
标签： #yyds干货盘点# LeetCode面试题：单词搜索 Java博客 51CTO博客
2023-04-17 18:23:53 246浏览
1.简述：
给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
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 1：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true
示例 2：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true
示例 3：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false
2.代码实现：
class Solution {
public boolean exist(char[][] board, String word) {
int h = board.length, w = board[0].length;
boolean[][] visited = new boolean[h][w];
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
boolean flag = check(board, visited, i, j, word, 0);
if (flag) {
return true;
}
}
}
return false;
}
public boolean check(char[][] board, boolean[][] visited, int i, int j, String s, int k) {
if (board[i][j] != s.charAt(k)) {
return false;
} else if (k == s.length() - 1) {
return true;
}
visited[i][j] = true;
int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
boolean result = false;
for (int[] dir : directions) {
int newi = i + dir[0], newj = j + dir[1];
if (newi >= 0 && newi < board.length && newj >= 0 && newj < board[0].length) {
if (!visited[newi][newj]) {
boolean flag = check(board, visited, newi, newj, s, k + 1);
if (flag) {
result = true;
break;
}
}
}
}
visited[i][j] = false;
return result;
}
}
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