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

参考文档