算法-单词搜索 II

1 题目概述

1.1 题目出处

https://leetcode.cn/problems/word-search-ii/description/?envType=study-plan-v2&envId=top-interview-150

1.2 题目描述

在这里插入图片描述

2 DFS

2.1 解题思路

每个格子往上下左右四个方向DFS，拼接后的单词如果在答案集中，则记录下来。

同时为了避免DFS时往回找，需要记录下已访问记录。

2.2 代码

class Solution {
    private Set<String> wordSet = new HashSet<>();
    private List<String> resultList = new LinkedList<>();

    public List<String> findWords(char[][] board, String[] words) {

        for (String word : words) {
            wordSet.add(word);
        }
        StringBuilder sb = new StringBuilder();
        char[][] visitSet = new char[board.length][board[0].length];

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                dfs(i, j, board, sb, visitSet);
            }
        }
        return resultList;
    }
    private void dfs(int i, int j, char[][] board, StringBuilder sb, char[][] visitSet) {
        if (sb.length() > 10) {
            return;
        }
        if (visitSet[i][j] == 1) {
            return;
        }
        visitSet[i][j] = 1;
        sb.append(board[i][j]);
        String currentStr = sb.toString();
        if (wordSet.contains(currentStr)) {
            resultList.add(currentStr);
            wordSet.remove(currentStr);
        }

        if (i > 0) {
            dfs(i - 1, j, board, sb, visitSet);
        }
        if (i < board.length - 1) {
            dfs(i + 1, j, board, sb, visitSet);
        }
        if (j > 0) {
            dfs(i, j - 1, board, sb, visitSet);
        }
        if (j < board[0].length - 1) {
            dfs(i, j + 1, board, sb, visitSet);
        }

        sb.deleteCharAt(sb.length() - 1);
        visitSet[i][j] = 0;
    }
}