标签：搜索 字典 efi item with rem info build rms
word squares you can build from them.A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns).
For example, the word sequence ["ball","area","lead","lady"] forms a word square because each word reads the same both horizontally and vertically.
b a l l
a r e a
l e a d
l a d y
a-z.Example 1:
Input:
["area","lead","wall","lady","ball"]
Output:
[["wall","area","lead","lady"],["ball","area","lead","lady"]]
Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
Example 2:
Input:
["abat","baba","atan","atal"]
Output:
[["baba","abat","baba","atan"],["baba","abat","baba","atal"]]
思路：字典树+dfs。
根据已添加单词确定下一个单词的前缀，利用dfs搜索。
public class Solution {
class TrieNode {
List<String> startWith;
TrieNode[] children;
TrieNode() {
startWith = new ArrayList<>();
children = new TrieNode[26];
}
}
class Trie {
TrieNode root;
Trie(String[] words) {
root = new TrieNode();
for (String w : words) {
TrieNode cur = root;
for (char ch : w.toCharArray()) {
int idx = ch - ‘a‘;
if (cur.children[idx] == null)
cur.children[idx] = new TrieNode();
cur.children[idx].startWith.add(w);
cur = cur.children[idx];
}
}
}
List<String> findByPrefix(String prefix) {
List<String> ans = new ArrayList<>();
TrieNode cur = root;
for (char ch : prefix.toCharArray()) {
int idx = ch - ‘a‘;
if (cur.children[idx] == null)
return ans;
cur = cur.children[idx];
}
ans.addAll(cur.startWith);
return ans;
}
}
public List<List<String>> wordSquares(String[] words) {
List<List<String>> ans = new ArrayList<>();
if (words == null || words.length == 0)
return ans;
int len = words[0].length();
Trie trie = new Trie(words);
List<String> ansBuilder = new ArrayList<>();
for (String w : words) {
ansBuilder.add(w);
search(len, trie, ans, ansBuilder);
ansBuilder.remove(ansBuilder.size() - 1);
}
return ans;
}
private void search(int len, Trie tr, List<List<String>> ans,
List<String> ansBuilder) {
if (ansBuilder.size() == len) {
ans.add(new ArrayList<>(ansBuilder));
return;
}
int idx = ansBuilder.size();
StringBuilder prefixBuilder = new StringBuilder();
for (String s : ansBuilder)
prefixBuilder.append(s.charAt(idx));
List<String> startWith = tr.findByPrefix(prefixBuilder.toString());
for (String sw : startWith) {
ansBuilder.add(sw);
search(len, tr, ans, ansBuilder);
ansBuilder.remove(ansBuilder.size() - 1);
}
}
}
标签：搜索 字典 efi item with rem info build rms
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078610.html
