Description
Given a
board which is a 2D matrix includes
a-z and dictionary
dict, find the largest collection of words on the board, the words can not overlap in the same position. return the
size of largest collection.
- The words in the dictionary are not repeated.
- You can reuse the words in the dictionary.
Example
Example 1:
Input:
["abc","def","ghi"]
{"abc","defi","gh"}
Output:
3
Explanation:
we can get the largest collection`["abc", "defi", "gh"]`
Example 2:
Input:
["aaaa","aaaa","aaaa","aaaa"]
{"a"}
Output:
16Explanation:
we can get the largest collection`["a", "a","a","a","a","a","a","a","a","a","a","a","a","a","a","a"]
思路：tire + dfs。
字典树用于前缀查找。
dfs用于搜索，
找到单词时搜索下一个单词
没有搜索到单词时，四方向遍历（回溯 + 标记）//建立tire树的过程
class Trie {
TrieNode root;
Trie() {
root = new TrieNode('0');
}
public void insert(String word) {
if(word == null || word.length() == 0) {
return;
}
TrieNode node = root;
for(int i = 0; i < word.length(); i++) {
char ch = word.charAt(i);
if(node.children[ch - 'a'] == null) {
node.children[ch - 'a'] = new TrieNode(ch);
}
node = node.children[ch - 'a'];
}
node.isWord = true;
}
}
//tire的结点
class TrieNode {
char value;
boolean isWord;
TrieNode[] children;
TrieNode(char v) {
value = v;
isWord = false;
children = new TrieNode[26];
}
}
public class Solution {
/**
* @param board a list of lists of character
* @param words a list of string
* @return an integer
*/
public int boggleGame(char[][] board, String[] words) {
// Write your code here
Trie trie = new Trie();
for(String word : words) {
trie.insert(word);
}
int m = board.length;
int n = board[0].length;
List<String> result = new ArrayList<>();
boolean[][] visited = new boolean[m][n];
List<String> path = new ArrayList<>();
findWords(result, board, visited, path, 0, 0, trie.root);
return result.size();
}
//从当前位置出发寻单词存不存在
public void findWords(List<String> result, char[][] board, boolean[][] visited, List<String> words, int x, int y, TrieNode root) {
int m = board.length;
int n = board[0].length;
for (int i = x; i < m; i++) {
for (int j = y; j < n; j++) {
List<List<Integer>> nextWordIndexes = new ArrayList<>();
List<Integer> path = new ArrayList<>();
getNextWords(nextWordIndexes, board, visited, path, i, j, root);
for (List<Integer> indexes : nextWordIndexes) {
String word = "";
for (int index : indexes) {
int row = index / n;
int col = index % n;
visited[row][col] = true;
word += board[row][col];
}
words.add(word);
if (words.size() > result.size()) {
result.clear();
result.addAll(words);
}
findWords(result, board, visited, words, i, j, root);
for (int index : indexes) {
int row = index / n;
int col = index % n;
visited[row][col] = false;
}
words.remove(words.size() - 1);
}
}
y = 0;
}
}
int []dx = {0, 1, 0, -1};
int []dy = {1, 0, -1, 0};
//dfs搜索查找单词
private void getNextWords(List<List<Integer>> words, char[][] board,
boolean[][] visited, List<Integer> path, int i, int j, TrieNode root) {
if(i < 0 | i >= board.length || j < 0 || j >= board[0].length
|| visited[i][j] == true || root.children[board[i][j] - 'a'] == null) {
return;
}
//找下一个单词
root = root.children[board[i][j] - 'a'];
if(root.isWord) {
List<Integer> newPath = new ArrayList<>(path);
newPath.add(i * board[0].length + j);
words.add(newPath);
return;
}
//回溯标记
visited[i][j] = true;
path.add(i * board[0].length + j);
for (int k = 0; k < 4; k ++) {
getNextWords(words, board, visited, path, i + dx[k], j + dy[k], root);
}
path.remove(path.size() - 1);
visited[i][j] = false;
}
}