Word Search II

Posted 2021-03-21 flagyuri

Given a matrix of lower alphabets and a dictionary. Find all words in the dictionary that can be found in the matrix. A word can start from any position in the matrix and go left/right/up/down to the adjacent position. One character only be used once in one word. No same word in dictionary

Example

Example 1:

Input：["doaf","agai","dcan"]，["dog","dad","dgdg","can","again"]
Output：["again","can","dad","dog"]
Explanation：
  d o a f
  a g a i
  d c a n
search in Matrix，so return ["again","can","dad","dog"].

Example 2:

Input：["a"]，["b"]
Output：[]
Explanation：
 a
search in Matrix，return [].

Challenge

Using trie to implement your algorithm.

 

使用了 Trie 的版本
考点：

• 字典树
• dfs
• tip:Tire树，一种树形结构，是一种哈希树的变种。典型应用是用于统计，排序和保存大量的字符串（但不仅限于字符串），所以经常被搜索引擎系统用于文本词频统计。

题解：

• 首先建立字典树，字典树从root开始，每个节点利用hashmap动态开点，利用字母的公共前缀建树。
• 遍历字母矩阵，将字母矩阵的每个字母，从root开始dfs搜索，搜索到底部时，将字符串存入答案返回即可。

  class TrieNode {    		//定义字典树的节点
      String word;
      HashMap<Character, TrieNode> children;   //使用HashMap动态开节点
      public TrieNode() {
          word = null;
          children = new HashMap<Character, TrieNode>();
      }
  };
  
  
  class TrieTree{
      TrieNode root;
      
      public TrieTree(TrieNode TrieNode) {
          root = TrieNode;
      }
      
      public void insert(String word) {		//字典树插入单词
          TrieNode node = root;
          for (int i = 0; i < word.length(); i++) {		
              if (!node.children.containsKey(word.charAt(i))) {
                  node.children.put(word.charAt(i), new TrieNode());
              }
              node = node.children.get(word.charAt(i));
          }
          node.word = word;
      }
  };
  
  public class Solution {
      /**
       * @param board: A list of lists of character
       * @param words: A list of string
       * @return: A list of string
       */
      public int[] dx = {1, 0, -1, 0};   //搜索方向
      public int[] dy = {0, 1, 0, -1};
      
      public void search(char[][] board,			//在字典树上dfs查找
                         int x,
                         int y,
                         TrieNode root,
                         List<String> results) {
          if (!root.children.containsKey(board[x][y])) {
              return;
          }
          
          TrieNode child = root.children.get(board[x][y]);
          
          if (child.word != null) {      //如果访问到字典树叶子，将字符串压入result即可
              if (!results.contains(child.word)) {
                  results.add(child.word);
              }
          }
          
          char tmp = board[x][y];
          board[x][y] = 0;  // mark board[x][y] as used
          
          for (int i = 0; i < 4; i++) {      //向四个方向dfs搜索
              if (!isValid(board, x + dx[i], y + dy[i])) {
                  continue;
              }
              search(board, x + dx[i], y + dy[i], child, results);
          }
          
          board[x][y] = tmp;  // revert the mark
      }
      
      private boolean isValid(char[][] board, int x, int y) {     //检测搜索位置合法
          if (x < 0 || x >= board.length || y < 0 || y >= board[0].length) {
              return false;
          }
          
          return board[x][y] != 0;
      }
      
      public List<String> wordSearchII(char[][] board, List<String> words) {
          List<String> results = new ArrayList<String>();
          
          TrieTree tree = new TrieTree(new TrieNode());
          for (String word : words){
              tree.insert(word);
          }
          
          for (int i = 0; i < board.length; i++) {				//遍历字母矩阵，将每个字母作为单词首字母开始搜索
              for (int j = 0; j < board[i].length; j++) {
                  search(board, i, j, tree.root, results);
              }
          }
          
          return results;
      }
  }