标签：cells use rdl ted row struct trie must private
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 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: board = [ [‘o‘,‘a‘,‘a‘,‘n‘], [‘e‘,‘t‘,‘a‘,‘e‘], [‘i‘,‘h‘,‘k‘,‘r‘], [‘i‘,‘f‘,‘l‘,‘v‘] ] words =["oath","pea","eat","rain"]Output:["eat","oath"]
Note:
a-z.words are distinct.Time: O(M * N *4^wordLen)
class Solution { int gRow; int gCol; public List<String> findWords(char[][] board, String[] words) { TrieNode root = build(words); gRow = board.length; gCol = board[0].length; List<String> res = new ArrayList<>(); for (int i = 0; i < gRow; i++) { for (int j = 0; j < gCol; j++) { // dfs(i, j, board, root, res); dfs(board, i, j, root, res); } } return res; } private void dfs(char[][] grid, int row, int col, TrieNode cur, List<String> res) { if (row < 0 || row >= gRow || col < 0 || col >= gCol) { return; } char c = grid[row][col]; if (c == ‘#‘ || cur.children[c - ‘a‘] == null) { return; } cur = cur.children[c - ‘a‘]; if (cur.word != null) { res.add(cur.word); cur.word = null; } grid[row][col] = ‘#‘; dfs(grid, row + 1, col, cur, res); dfs(grid, row - 1, col, cur, res); dfs(grid, row, col + 1, cur, res); dfs(grid, row, col - 1, cur, res); grid[row][col] = c; } private TrieNode build(String[] words) { TrieNode root = new TrieNode(); for (String word : words) { TrieNode cur = root; char[] charArr = word.toCharArray(); for (char c : charArr) { if (cur.children[c - ‘a‘] == null) { cur.children[c - ‘a‘] = new TrieNode(); } cur = cur.children[c - ‘a‘]; } cur.word = word; } return root; } } class TrieNode { TrieNode[] children; String word; public TrieNode() { children = new TrieNode[26]; word = null; } }
标签：cells use rdl ted row struct trie must private
原文地址：https://www.cnblogs.com/xuanlu/p/12710903.html
