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.
[思路]
1. 按照word search I的思路, 超时.
2. trie 前缀树, 可以pruning 剪枝. pass.
[CODE]
1. 超时
public class Solution {public List<String> findWords(char[][] board, String[] words) {List<String> res = new ArrayList<String>();if(board==null || words==null || board.length==0 || words.length==0) return res;boolean[][] visited = new boolean[board.length][board[0].length];Set<String> dict = new HashSet<String>(Arrays.asList(words));for(int i=0; i<board.length; i++) {for(int j=0; j<board[0].length; j++) {search(board, visited, dict, i, j, new StringBuilder(), res);}}return res;}private void search(char[][] board,boolean[][] visited,Set<String> dict,int i,int j, StringBuilder sb, List<String> res) {if(i<0 || i>board.length-1 || j<0 || j>board[0].length-1 || visited[i][j]) return;sb.append(board[i][j]);visited[i][j] = true;if(dict.contains(sb.toString()))res.add(sb.toString());search(board, visited, dict, i-1, j, sb, res);search(board, visited, dict, i+1, j, sb, res);search(board, visited, dict, i, j-1, sb, res);search(board, visited, dict, i, j+1, sb, res);sb.deleteCharAt(sb.length() – 1);visited[i][j] = false;}}2. 使用trie, 检查前缀, pruningpublic class Solution {public List<String> findWords(char[][] board, String[] words) {Set<String> res = new HashSet<String>();if(board==null || words==null || board.length==0 || words.length==0) return new ArrayList<String>(res);boolean[][] visited = new boolean[board.length][board[0].length];Trie trie = new Trie();for(String word : words) {trie.insert(word);}for(int i=0; i<board.length; i++) {for(int j=0; j<board[0].length; j++) {search(board, visited, trie, i, j, new StringBuilder(), res);}}return new ArrayList<String>(res);}private void search(char[][] board,boolean[][] visited,Trie trie,int i,int j, StringBuilder sb, Set<String> res) {if(i<0 || i>board.length-1 || j<0 || j>board[0].length-1 || visited[i][j]) return;sb.append(board[i][j]);String s = sb.toString();visited[i][j] = true;if(trie.startsWith(s)) {if(trie.search(s)) res.add(s);search(board, visited, trie, i-1, j, sb, res);search(board, visited, trie, i+1, j, sb, res);search(board, visited, trie, i, j-1, sb, res);search(board, visited, trie, i, j+1, sb, res);}sb.deleteCharAt(sb.length() – 1);visited[i][j] = false;}}class TrieNode {// Initialize your data structure here.char c;boolean leaf;HashMap<Character, TrieNode> children = new HashMap<Character, TrieNode>();public TrieNode(char c) {this.c = c;}public TrieNode(){};}class Trie {private TrieNode root;public Trie() {root = new TrieNode();}// Inserts a word into the trie.public void insert(String word) {Map<Character, TrieNode> children = root.children;for(int i=0; i<word.length(); i++) {char c = word.charAt(i);TrieNode t;if(children.containsKey(c)) {t = children.get(c);} else {t = new TrieNode(c);children.put(c, t);}children = t.children;if(i==word.length()-1) t.leaf=true;}}// Returns if the word is in the trie.public boolean search(String word) {TrieNode t = searchNode(word);return t!=null && t.leaf;}public boolean startsWith(String prefix) {return searchNode(prefix) != null;}private TrieNode searchNode(String word) {Map<Character, TrieNode> children = root.children;TrieNode t = null;for(int i=0; i<word.length(); i++) {char c = word.charAt(i);if(!children.containsKey(c)) return null;t = children.get(c);children = t.children;}return t;}}
，第一个青春是上帝给的;第二个的青春是靠自己努力的
