算法-单词搜索 II
算法-单词搜索 II
1 题目概述
1.1 题目出处
https://leetcode.cn/problems/word-search-ii/description/?envType=study-plan-v2&envId=top-interview-150
1.2 题目描述
2 DFS
2.1 解题思路
每个格子往上下左右四个方向DFS，拼接后的单词如果在答案集中，则记录下来。
同时为了避免DFS时往回找，需要记录下已访问记录。
2.2 代码
class Solution {
private Set<String> wordSet = new HashSet<>();
private List<String> resultList = new LinkedList<>();
public List<String> findWords(char[][] board, String[] words) {
for (String word : words) {
wordSet.add(word);
}
StringBuilder sb = new StringBuilder();
char[][] visitSet = new char[board.length][board[0].length];
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
dfs(i, j, board, sb, visitSet);
}
}
return resultList;
}
private void dfs(int i, int j, char[][] board, StringBuilder sb, char[][] visitSet) {
if (sb.length() > 10) {
return;
}
if (visitSet[i][j] == 1) {
return;
}
visitSet[i][j] = 1;
sb.append(board[i][j]);
String currentStr = sb.toString();
if (wordSet.contains(currentStr)) {
resultList.add(currentStr);
wordSet.remove(currentStr);
}
if (i > 0) {
dfs(i - 1, j, board, sb, visitSet);
}
if (i < board.length - 1) {
dfs(i + 1, j, board, sb, visitSet);
}
if (j > 0) {
dfs(i, j - 1, board, sb, visitSet);
}
if (j < board[0].length - 1) {
dfs(i, j + 1, board, sb, visitSet);
}
sb.deleteCharAt(sb.length() - 1);
visitSet[i][j] = 0;
}
}
2.3 时间复杂度
O(M * N * 4^10) 字符串最多10
2.4 空间复杂度
O(10)
3 DFS+Trie树
3.1 解题思路
3.2 代码
class Solution {
private Set<String> wordSet = new HashSet<>();
private List<String> resultList = new LinkedList<>();
public List<String> findWords(char[][] board, String[] words) {
for (String word : words) {
wordSet.add(word);
}
StringBuilder sb = new StringBuilder();
char[][] visitSet = new char[board.length][board[0].length];
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
dfs(i, j, board, sb, visitSet);
}
}
return resultList;
}
private void dfs(int i, int j, char[][] board, StringBuilder sb, char[][] visitSet) {
if (sb.length() > 10) {
return;
}
if (visitSet[i][j] == 1) {
return;
}
visitSet[i][j] = 1;
sb.append(board[i][j]);
String currentStr = sb.toString();
if (wordSet.contains(currentStr)) {
resultList.add(currentStr);
wordSet.remove(currentStr);
}
if (i > 0) {
dfs(i - 1, j, board, sb, visitSet);
}
if (i < board.length - 1) {
dfs(i + 1, j, board, sb, visitSet);
}
if (j > 0) {
dfs(i, j - 1, board, sb, visitSet);
}
if (j < board[0].length - 1) {
dfs(i, j + 1, board, sb, visitSet);
}
sb.deleteCharAt(sb.length() - 1);
visitSet[i][j] = 0;
}
}
3.3 时间复杂度
4 DFS+Trie树 优化
4.1 解题思路
4.2 代码
class Solution {
private List<String> resultList = new LinkedList<>();
private TrieNode trieNode = new TrieNode();
static class TrieNode {
private TrieNode[] trieNodes = new TrieNode[26];
public boolean isWord = false;
public void insert(String word) {
if (word.length() == 0) {
isWord = true;
return;
}
int index = word.charAt(0) - 'a';
if (null == trieNodes[index]) {
trieNodes[index] = new TrieNode();
}
trieNodes[index].insert(word.substring(1));
}
}
public List<String> findWords(char[][] board, String[] words) {
for (String word : words) {
trieNode.insert(word);
}
StringBuilder sb = new StringBuilder();
char[][] visitSet = new char[board.length][board[0].length];
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
dfs(i, j, board, sb, visitSet, trieNode);
}
}
return resultList;
}
private void dfs(int i, int j, char[][] board, StringBuilder sb, char[][] visitSet, TrieNode ct) {
if (sb.length() > 10) {
return;
}
if (visitSet[i][j] == 1) {
return;
}
visitSet[i][j] = 1;
sb.append(board[i][j]);
ct = ct.trieNodes[board[i][j] - 'a'];
if (null != ct) {
if (ct.isWord) {
resultList.add(sb.toString());
ct.isWord = false;
}
if (i > 0) {
dfs(i - 1, j, board, sb, visitSet, ct);
}
if (i < board.length - 1) {
dfs(i + 1, j, board, sb, visitSet, ct);
}
if (j > 0) {
dfs(i, j - 1, board, sb, visitSet, ct);
}
if (j < board[0].length - 1) {
dfs(i, j + 1, board, sb, visitSet, ct);
}
}
sb.deleteCharAt(sb.length() - 1);
visitSet[i][j] = 0;
}
}
4.3 时间复杂度
