LeetCode 刷题之路 Word Search II

最新推荐文章于 2020-02-15 12:37:41 发布

JianS_Ted

最新推荐文章于 2020-02-15 12:37:41 发布

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分类专栏： Java 刷题记录 LeetCode 文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/JianS_Ted/article/details/80860405

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本文介绍了一种使用Trie树和深度优先搜索(DFS)解决二维板搜索词问题的方法。通过构建Trie树来存储词典中的所有单词，并利用DFS遍历二维板，找到所有可能形成的词。文章详细阐述了算法实现细节及示例。

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Word Search II

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of the sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example:

Input: 
words = ["oath","pea","eat","rain"] and board =
[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Output: ["eat","oath"]

Note:
You may assume that all inputs are consist of lowercase letters a-z.

这题是在Word Search 基础上拓展而来的。很明显我们需要用到DFS来进行搜索。对于这种找某个string是否存在，Trie是个很有力的工具(因为查找检索快捷)，所以这题采用了Trie+DFS的方法进行求解。

这里利用w.toCharArray()代替w.charAt(i) 以及在Tire树里面添加一个String变量来便于后期结果获取(这样就可以避免StringBuilder使用上的繁琐)。

class Solution {
    class TireNode{
        public TireNode[] children;
        public String word;
        public TireNode(){
            children = new TireNode[26];
            word = null;
        }
    }

    public TireNode tireBuilder(String[] words){
        TireNode root = new TireNode();
        for(String w : words){
            TireNode input = root;
            for(char cur : w.toCharArray()){
                int index = cur - 'a';
                if(input.children[index] == null){
                    input.children[index] = new TireNode();
                }
                input = input.children[index];
            }
            input.word = w;
        }
        return root;
    }

    public void dfs(char[][] board, int i, int j, TireNode start, List<String> result){
        if(i<0 || i== board.length
                || j<0 || j==board[0].length){
            return;
        }

        char cur = board[i][j];
        // Stop searching when the char has been visited or the word has been found
        if(cur == '!' || start.children[cur-'a']==null){
            return;
        }
        start = start.children[cur-'a'];
        if(start.word!=null){
            result.add(start.word);
            start.word = null;
        }

        board[i][j] = '!';
        dfs(board, i+1, j, start, result);
        dfs(board, i-1, j, start, result);
        dfs(board, i, j+1, start, result);
        dfs(board, i, j-1, start, result);
        
        board[i][j] = cur;

    }

    public List<String> findWords(char[][] board, String[] words) {
        List<String> result = new ArrayList<>();
        TireNode root = this.tireBuilder(words);
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                dfs(board, i, j, root, result);
            }
        }
        return result;
    }
}