113 lines
3.3 KiB
Java
113 lines
3.3 KiB
Java
//给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words，找出所有同时在二维网格和字典中出现的单词。
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//
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// 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在一个单词中不允许被重复使
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//用。
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f",
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//"l","v"]], words = ["oath","pea","eat","rain"]
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//输出：["eat","oath"]
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//
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//
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// 示例 2：
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//
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//
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//输入：board = [["a","b"],["c","d"]], words = ["abcb"]
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//输出：[]
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//
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//
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//
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//
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// 提示：
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//
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//
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// m == board.length
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// n == board[i].length
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// 1 <= m, n <= 12
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// board[i][j] 是一个小写英文字母
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// 1 <= words.length <= 3 * 10⁴
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// 1 <= words[i].length <= 10
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// words[i] 由小写英文字母组成
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// words 中的所有字符串互不相同
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//
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// Related Topics 字典树 数组 字符串 回溯 矩阵 👍 531 👎 0
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package leetcode.editor.cn;
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import java.util.*;
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//212:单词搜索 II
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class WordSearchIi {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new WordSearchIi().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
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public List<String> findWords(char[][] board, String[] words) {
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Trie trie = new Trie();
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for (String word : words) {
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trie.insert(word);
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}
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Set<String> ans = new HashSet<String>();
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for (int i = 0; i < board.length; ++i) {
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for (int j = 0; j < board[0].length; ++j) {
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dfs(board, trie, i, j, ans);
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}
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}
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return new ArrayList<String>(ans);
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}
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public void dfs(char[][] board, Trie now, int i1, int j1, Set<String> ans) {
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if (!now.children.containsKey(board[i1][j1])) {
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return;
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}
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char ch = board[i1][j1];
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now = now.children.get(ch);
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if (!"".equals(now.word)) {
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ans.add(now.word);
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}
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board[i1][j1] = '#';
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for (int[] dir : dirs) {
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int i2 = i1 + dir[0], j2 = j1 + dir[1];
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if (i2 >= 0 && i2 < board.length && j2 >= 0 && j2 < board[0].length) {
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dfs(board, now, i2, j2, ans);
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}
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}
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board[i1][j1] = ch;
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}
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}
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class Trie {
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String word;
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Map<Character, Trie> children;
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boolean isWord;
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public Trie() {
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this.word = "";
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this.children = new HashMap<Character, Trie>();
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}
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public void insert(String word) {
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Trie cur = this;
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for (int i = 0; i < word.length(); ++i) {
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char c = word.charAt(i);
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if (!cur.children.containsKey(c)) {
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cur.children.put(c, new Trie());
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}
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cur = cur.children.get(c);
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}
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cur.word = word;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |