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;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51

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;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51

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;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73

4.3 时间复杂度

参考文档