題目

https://leetcode.com/problems/word-search-ii/

題解

基于前綴樹實現，如果把 Trie 也看做一個特殊的圖的話，就是 將兩個圖同時進行 dfs，就像判斷兩個樹是否相同那樣。

可以參考：leetcode 208. Implement Trie (Prefix Tree) | 208. 實現 Trie 前綴樹（Java） 中的數據結構，然后改造一下即可。

class Node {Node[] map;boolean isEnd;public Node() {this.map = new Node['z' - 'a' + 1];}public Node get(char c) {return map[c - 'a'];}public Node put(char c) {Node node = new Node();map[c - 'a'] = node;return node;} }class Trie {Node node;public Trie() {node = new Node();}public void insert(String word) {char[] chars = word.toCharArray();Node cur = node;for (int i = 0; i < chars.length; i++) {if (cur.get(chars[i]) != null) {cur = cur.get(chars[i]);} else {cur = cur.put(chars[i]);}if (i == chars.length - 1) {cur.isEnd = true;}}} }class Solution {int M, N;List<String> result;public List<String> findWords(char[][] board, String[] words) {M = board.length;N = board[0].length;result = new ArrayList<>();Trie trie = new Trie();for (String s : words) {trie.insert(s);}boolean[][] visited = new boolean[M][N];for (int k = 0; k < trie.node.map.length; k++) {if (trie.node.map[k] == null) continue;// word can start from any location on boardfor (int i = 0; i < M; i++) {for (int j = 0; j < N; j++) {dfs(board, i, j, visited, trie.node.map[k], (char) (k + 'a'), String.valueOf((char) (k + 'a')));}}}return result;}public void dfs(char[][] board, int i, int j, boolean[][] visited, Node node, char val, String path) {if (i < 0 || i == M || j < 0 || j == N || visited[i][j] || board[i][j] != val) return;if (node.isEnd) {result.add(path);node.isEnd = false; // 避免相同前綴的字符串被多次添加到result中}visited[i][j] = true;for (int k = 0; k < node.map.length; k++) {Node next = node.map[k];if (next != null) {dfs(board, i + 1, j, visited, next, (char) (k + 'a'), path + ((char) (k + 'a')));dfs(board, i - 1, j, visited, next, (char) (k + 'a'), path + ((char) (k + 'a')));dfs(board, i, j + 1, visited, next, (char) (k + 'a'), path + ((char) (k + 'a')));dfs(board, i, j - 1, visited, next, (char) (k + 'a'), path + ((char) (k + 'a')));}}visited[i][j] = false;} }

總結

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