Valid Word Square
题目链接:https://leetcode.com/problems...
暴力遍历,一个一个检查看符不符合要求。
public boolean validWordSquare(List<String> words) {
/* words[i][j] == words[j][i]
* j >= len(words) or i >= len(words[j]) return false
*/
for(int i = 0; i < words.size(); i++) {
String word = words.get(i);
for(int j = 0; j < word.length(); j++) {
if(j >= words.size() || i >= words.get(j).length()) return false;
if(word.charAt(j) != words.get(j).charAt(i)) return false;
}
}
return true;
}
Word Squares
题目链接:https://leetcode.com/problems...
这道题如果用上一题的方法,一个一个试的话,时间复杂度太高。所以要另想办法。
首先这种需要求出所有结果的题,一般都是用dfs(backtracking)的。然后这道题的思路是单词一个一个确定。因为题目已经说了word的长度范围是1到5,最多考虑五个单词即可。
第一个单词:""为前缀,在words数组里随便取一个word1,确定长度
第二个单词:在剩下的words里取出一个以word1[1]为前缀的word2
第三个单词:在剩下的里取出一个以word1[2]+word2[2]为前缀的word3
第四个单词:在剩下的里取出一个以word1[3]+word2[3]+word3[3]为前缀的word4
第五个单词:在剩下的里取出一个以word1[4]+word2[4]+word3[4]+word4[4]为前缀的word5
所以这题需要快速的找到前缀,那么可以想到用hashmap或者trie tree。题目说了单词没有duplication,省去了查重的过程。
遍历words,把其中的一个单词当作1st word
找到第二个单词,加到square里面,接着找第三个单词......这是个backtracking的过程,如果prefix不在trie tree里面直接return
找第二个,第三个,。。。单词的过程用的是bfs,已经知道prefix之后,根据prefix找到trie tree里面对应的node,从改node开始bfs走剩下的长度,找到所有可能的node,检查这些node是否是单词的末尾,是的话就放入list里面,给上面dfs的method来用
public class Solution {
public List<List<String>> wordSquares(String[] words) {
// hashmap or trie tree
TrieNode root = buildTrieTree(words);
// traverse all word int words, for the 1st word
for(String word : words) {
// len(word) == 1, itself is square
if(word.length() == 1) {
List<String> cur = new ArrayList();
cur.add(word);
result.add(cur);
continue;
}
List<String> square = new ArrayList();
square.add(word);
dfs(root, square);
}
return result;
}
List<List<String>> result = new ArrayList();
/* from the idx position in words
* 1st word: a = words[i]
* 2nd word: b prefix = a.charAt(1)
* 3rd word: c prefix = a.charAt(2) + b.charAt(2)
* 4th word: d prefix = a.charAt(3) + b.charAt(3) + c.charAt(3)
* 5th word: e prefix = a.charAt(4) + b.charAt(4) + c.charAt(4) + d.charAt(4)
*/
private void dfs(TrieNode root, List<String> square) {
int len = square.get(0).length();
int prefix_length = square.size();
if(len == prefix_length) {
result.add(new ArrayList(square));
return;
}
String prefix = "";
// find supposed prefix
for(int i = 0; i < prefix_length; i++) prefix += square.get(i).charAt(prefix_length);
// find word with that prefix in trie tree
TrieNode node = root;
for(int i = 0; i < prefix.length(); i++) {
if(node.children[getIndex(prefix.charAt(i))] == null) return;
node = node.children[getIndex(prefix.charAt(i))];
}
List<String> next = bfs(node, len - prefix_length);
if(next.size() == 0) return;
for(String s : next) {
square.add(s);
dfs(root, square);
square.remove(square.size() - 1);
}
}
// find all possible words with certain prefix
private List<String> bfs(TrieNode node, int left) {
List<String> possible = new ArrayList();
Queue<TrieNode> q = new LinkedList();
q.offer(node);
while(left-- > 0) {
if(q.isEmpty()) break;
for(int j = q.size(); j > 0; j--) {
TrieNode cur = q.poll();
for(int i = 0; i < 26; i++) {
if(cur.children[i] != null) q.offer(cur.children[i]);
}
}
}
for(TrieNode cur : q) {
if(cur.word != null) possible.add(cur.word);
}
return possible;
}
private int getIndex(char c) {
return c - 'a';
}
class TrieNode {
TrieNode[] children = new TrieNode[26];
String word = "";
}
private TrieNode buildTrieTree(String[] words) {
TrieNode root = new TrieNode();
for(String word : words) {
TrieNode cur = root;
for(int i = 0; i < word.length(); i++) {
if(cur.children[getIndex(word.charAt(i))] == null) {
cur.children[getIndex(word.charAt(i))] = new TrieNode();
}
cur = cur.children[getIndex(word.charAt(i))];
}
cur.word = word;
}
return root;
}
}
看了discussion的方法,直接在构造trie的时候,在node里面加上以它为prefix所有可能的word,这样多加了一个List<String>的空间,但是省去了上面bfs找单词的时间。
public class Solution {
public List<List<String>> wordSquares(String[] words) {
// hashmap or trie tree
TrieNode root = buildTrieTree(words);
// traverse all word int words, for the 1st word
for(String word : words) {
// len(word) == 1, itself is square
if(word.length() == 1) {
List<String> cur = new ArrayList();
cur.add(word);
result.add(cur);
continue;
}
List<String> square = new ArrayList();
square.add(word);
dfs(root, square);
}
return result;
}
List<List<String>> result = new ArrayList();
private void dfs(TrieNode root, List<String> square) {
int len = square.get(0).length();
if(len == square.size()) {
result.add(new ArrayList(square));
return;
}
// find all words with prefix
List<String> next = findWords(root, square);
for(String s : next) {
square.add(s);
dfs(root, square);
square.remove(square.size() - 1);
}
}
private List<String> findWords(TrieNode node, List<String> square) {
String prefix = "";
int prefix_length = square.size();
// find supposed prefix
for(int i = 0; i < prefix_length; i++) prefix += square.get(i).charAt(prefix_length);
// find word with that prefix in trie tree
for(int i = 0; i < prefix.length(); i++) {
if(node.children[getIndex(prefix.charAt(i))] == null) return new ArrayList();
node = node.children[getIndex(prefix.charAt(i))];
}
return node.wordsWithPrefix;
}
private int getIndex(char c) {
return c - 'a';
}
class TrieNode {
TrieNode[] children = new TrieNode[26];
List<String> wordsWithPrefix = new ArrayList();
}
private TrieNode buildTrieTree(String[] words) {
TrieNode root = new TrieNode();
for(String word : words) {
TrieNode cur = root;
for(int i = 0; i < word.length(); i++) {
if(cur.children[getIndex(word.charAt(i))] == null) {
cur.children[getIndex(word.charAt(i))] = new TrieNode();
}
cur = cur.children[getIndex(word.charAt(i))];
cur.wordsWithPrefix.add(word);
}
}
return root;
}
}
Trie Tree的构造方法
cc150上面给了Trie和TrieNode的构造方法。感觉lc里面主要是高清TrieNode的fields,以及对应的build一个Trie的方法。lc上一共7道trie的题,出现过的TrieNode的fields不同的选择好像就3种。
首先children是肯定都需要的,两种:array(TrieNode[])或者HashMap(Map<Character, TrieNode>)。都可以用,但是好像lc上的题基本用的都是array,character的范围比较小用array还是比较省空间。
然后还用到的fields有:isWord(boolean:结尾,判断是否形成一个单词), word(String:结尾,形成的单词)以及startsWith(List<String>:以走到当前的TrieNode形成的String为prefix,所有的word)
如果题目只要求判断单词在不在字典里,isWord就够了。
如果题目要求返回String,那么一般用word。
如果题目要求返回所有以特定的prefix开头的单词,那么可以用startsWith。
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