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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.
For example,
Given words = ["oath","pea","eat","rain"] and board =
[
[‘o‘,‘a‘,‘a‘,‘n‘],
[‘e‘,‘t‘,‘a‘,‘e‘],
[‘i‘,‘h‘,‘k‘,‘r‘],
[‘i‘,‘f‘,‘l‘,‘v‘]
]
Return ["eat","oath"].
Note:
You may assume that all inputs are consist of lowercase letters a-z.
You would need to optimize your backtracking to pass the larger test. Could you stop backtracking earlier?
If the current candidate does not exist in all words‘ prefix, you could stop backtracking immediately. What kind of data structure could answer such query efficiently? Does a hash table work? Why or why not? How about a Trie? If you would like to learn how to implement a basic trie, please work on this problem: Implement Trie (Prefix Tree) first.
public class Solution { public class TrieNode{ TrieNode[] childs = new TrieNode[26]; String word; } //int[][] moves = new int[][] { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } }; public List<String> findWords(char[][] board, String[] words) { List<String> resList = new ArrayList<String>(); if (board.length==0 || board[0].length==0) return resList; TrieNode root = new TrieNode(); for (String word : words){ addWord(root,word.toCharArray()); } for (int i=0;i<board.length;i++) for (int j=0;j<board[0].length;j++) if (root.childs[board[i][j]-‘a‘]!=null){ findWordsRecur(board,root.childs[board[i][j]-‘a‘],i,j,resList); } return resList; } // Assume that board[x][y] matches curNode, starting from [x,y] search for other words whose root is curNode. public void findWordsRecur(char[][] board, TrieNode curNode, int x, int y, List<String> resList){ // Find a word. if (curNode.word!=null){ resList.add(curNode.word); // OPTIMIZATION: Set word to null for one-time add, instead of using HashSet for duplicates. curNode.word = null; } // OPTIMIZATION: Mark board[x][y] as ‘#‘ instead of using boolean[][] visited. char curChar = board[x][y]; board[x][y] = ‘#‘; // OPTIMIZATION: Use four explicit conditions instead of loop and isValid function. if (x>0 && board[x-1][y]!=‘#‘ && curNode.childs[board[x-1][y]-‘a‘]!=null) findWordsRecur(board,curNode.childs[board[x-1][y]-‘a‘],x-1,y,resList); if (x<board.length-1 && board[x+1][y]!=‘#‘ && curNode.childs[board[x+1][y]-‘a‘]!=null) findWordsRecur(board,curNode.childs[board[x+1][y]-‘a‘],x+1,y,resList); if (y>0 && board[x][y-1]!=‘#‘ && curNode.childs[board[x][y-1]-‘a‘]!=null) findWordsRecur(board,curNode.childs[board[x][y-1]-‘a‘],x,y-1,resList); if (y<board[0].length-1 && board[x][y+1]!=‘#‘ && curNode.childs[board[x][y+1]-‘a‘]!=null) findWordsRecur(board,curNode.childs[board[x][y+1]-‘a‘],x,y+1,resList); board[x][y] = curChar; } /*public boolean isValid(char[][] board, int x, int y){ return x>=0 && x<board.length && y>=0 && y<board[0].length; }*/ public void addWord(TrieNode root, char[] word){ for (int i=0;i<word.length;i++){ char c = word[i]; if (root.childs[c-‘a‘]==null){ root.childs[c-‘a‘] = new TrieNode(); } root = root.childs[c-‘a‘]; } root.word = new StringBuilder().append(word).toString(); } }
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原文地址：http://www.cnblogs.com/lishiblog/p/5858639.html
