#yyds干货盘点# LeetCode程序员面试金典：单词搜索 II

1.简述：

给定一个m x n 二维字符网格board和单词(字符串)列表 words，在二维网格上返回所有单词。

单词必须按字母顺序通过 相邻单元格 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单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

示例 1：

输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]

输出：["eat","oath"]

示例 2：

输入：board = [["a","b"],["c","d"]], words = ["abcb"]

输出：[]

2.实现代码：

class Solution {    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};    public List<String> findWords(char[][] board, String[] words) {        Trie trie = new Trie();        for (String word : words) {            trie.insert(word);        }        Set<String> ans = new HashSet<String>();        for (int i = 0; i < board.length; ++i) {            for (int j = 0; j < board[0].length; ++j) {                dfs(board, trie, i, j, ans);            }        }        return new ArrayList<String>(ans);    }    public void dfs(char[][] board, Trie now, int i1, int j1, Set<String> ans) {        if (!now.children.containsKey(board[i1][j1]) {            return;        }        char ch = board[i1][j1];        now = now.children.get(ch);        if (!"".equals(now.word)) {            ans.add(now.word);        }        board[i1][j1] = '#';        for (int[] dir : dirs) {            int i2 = i1 + dir[0], j2 = j1 + dir[1];            if (i2 >= 0 && i2 < board.length && j2 >= 0 && j2 < board[0].length) {                dfs(board, now, i2, j2, ans);            }        }        board[i1][j1] = ch;    }}class Trie {    String word;    Map<Character, Trie> children;    boolean isWord;    public Trie() {        this.word = "";        this.children = new HashMap<Character, Trie>();    }    public void insert(String word) {        Trie cur = this;        for (int i = 0; i < word.length(); ++i) {            char c = word.charAt(i);            if (!cur.children.containsKey(c)) {                cur.children.put(c, new Trie());            }            cur = cur.children.get(c);        }        cur.word = word;    }}