给定字典，这是一种在字典和M x N板中进行查询的方法，其中每个单元格都有一个字符。查找可以由一系列相邻字符组成的所有可能单词。请注意，我们可以移至8个相邻字符的任何一个，但是一个单词不应具有同一单元格的多个实例。
例子：
输入： dictionary [] = {“ GEEKS”，“ FOR”，“ QUIZ”，“ GO”}; boggle [] [] = {{‘G’，’I’，’Z’}，{‘U’，’E’，’K’}，{‘Q’，’S’，’E’}};输出：出现词典的以下单词GEEKS QUIZ说明： _0.jpg)
输入： dictionary [] = {“ GEEKS”，“ ABCFIHGDE”}; boggle [] [] = {{‘A’，’B’，’C’}，{‘D’，’E’，’F’}，{‘G’，’H’，’I’}};输出：字典中以下单词出现ABCFIHGDE说明： _1.jpg)
我们在下面的文章中讨论了基于Graph DFS的解决方案。
Boggle(在一个字符板上查找所有可能的单词)|套装1
在这里，我们讨论基于Trie的解决方案，它比基于DFS的解决方案更好。
给定字典dictionary [] = {“ GEEKS”，“ FOR”，“ QUIZ”，“ GO”}
1.创建一个空的trie并将给定字典的所有单词插入trie
After insertion, Trie looks like(leaf nodes are in RED)
root
/
G F Q
/ | | |
O E O U
| | |
E R I
| |
K Z
|
S
2.之后，我们仅在boggle [] []中选择那些是Trie根的子代的字符
在上面让我们选择’G’boggle [0] [0]，’Q’boggle [2] [0](它们都存在于boggle矩阵中)
3.在trie中搜索一个以我们在第2步中选择的字符开头的单词
1) Create bool visited boolean matrix (Visited[M][N] = false )
2) Call SearchWord() for every cell (i, j) which has one of the
first characters of dictionary words. In above example,
we have 'G' and 'Q' as first characters.
SearchWord(Trie *root, i, j, visited[][N])
if root->leaf == true
print word
if we have seen this element first time then make it visited.
visited[i][j] = true
do
traverse all child of current root
k goes (0 to 26 ) [there are only 26 Alphabet]
add current char and search for next character
find next character which is adjacent to boggle[i][j]
they are 8 adjacent cells of boggle[i][j] (i+1, j+1),
(i+1, j) (i-1, j) and so on.
make it unvisited visited[i][j] = false
下面是上述想法的实现：
C++
// C++ program for Boggle game
#include
using namespace std;
// Converts key current character into index
// use only 'A' through 'Z'
#define char_int(c) ((int)c - (int)'A')
// Alphabet size
#define SIZE (26)
#define M 3
#define N 3
// trie Node
struct TrieNode {
TrieNode* Child[SIZE];
// isLeaf is true if the node represents
// end of a word
bool leaf;
};
// Returns new trie node (initialized to NULLs)
TrieNode* getNode()
{
TrieNode* newNode = new TrieNode;
newNode->leaf = false;
for (int i = 0; i < SIZE; i++)
newNode->Child[i] = NULL;
return newNode;
}
// If not present, inserts a key into the trie
// If the key is a prefix of trie node, just
// marks leaf node
void insert(TrieNode* root, char* Key)
{
int n = strlen(Key);
TrieNode* pChild = root;
for (int i = 0; i < n; i++) {
int index = char_int(Key[i]);
if (pChild->Child[index] == NULL)
pChild->Child[index] = getNode();
pChild = pChild->Child[index];
}
// make last node as leaf node
pChild->leaf = true;
}
// function to check that current location
// (i and j) is in matrix range
bool isSafe(int i, int j, bool visited[M][N])
{
return (i >= 0 && i < M && j >= 0 && j < N && !visited[i][j]);
}
// A recursive function to print all words present on boggle
void searchWord(TrieNode* root, char boggle[M][N], int i,
int j, bool visited[][N], string str)
{
// if we found word in trie / dictionary
if (root->leaf == true)
cout << str << endl;
// If both I and j in range and we visited
// that element of matrix first time
if (isSafe(i, j, visited)) {
// make it visited
visited[i][j] = true;
// traverse all childs of current root
for (int K = 0; K < SIZE; K++) {
if (root->Child[K] != NULL) {
// current character
char ch = (char)K + (char)'A';
// Recursively search reaming character of word
// in trie for all 8 adjacent cells of boggle[i][j]
if (isSafe(i + 1, j + 1, visited)
&& boggle[i + 1][j + 1] == ch)
searchWord(root->Child[K], boggle,
i + 1, j + 1, visited, str + ch);
if (isSafe(i, j + 1, visited)
&& boggle[i][j + 1] == ch)
searchWord(root->Child[K], boggle,
i, j + 1, visited, str + ch);
if (isSafe(i - 1, j + 1, visited)
&& boggle[i - 1][j + 1] == ch)
searchWord(root->Child[K], boggle,
i - 1, j + 1, visited, str + ch);
if (isSafe(i + 1, j, visited)
&& boggle[i + 1][j] == ch)
searchWord(root->Child[K], boggle,
i + 1, j, visited, str + ch);
if (isSafe(i + 1, j - 1, visited)
&& boggle[i + 1][j - 1] == ch)
searchWord(root->Child[K], boggle,
i + 1, j - 1, visited, str + ch);
if (isSafe(i, j - 1, visited)
&& boggle[i][j - 1] == ch)
searchWord(root->Child[K], boggle,
i, j - 1, visited, str + ch);
if (isSafe(i - 1, j - 1, visited)
&& boggle[i - 1][j - 1] == ch)
searchWord(root->Child[K], boggle,
i - 1, j - 1, visited, str + ch);
if (isSafe(i - 1, j, visited)
&& boggle[i - 1][j] == ch)
searchWord(root->Child[K], boggle,
i - 1, j, visited, str + ch);
}
}
// make current element unvisited
visited[i][j] = false;
}
}
// Prints all words present in dictionary.
void findWords(char boggle[M][N], TrieNode* root)
{
// Mark all characters as not visited
bool visited[M][N];
memset(visited, false, sizeof(visited));
TrieNode* pChild = root;
string str = "";
// traverse all matrix elements
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// we start searching for word in dictionary
// if we found a character which is child
// of Trie root
if (pChild->Child[char_int(boggle[i][j])]) {
str = str + boggle[i][j];
searchWord(pChild->Child[char_int(boggle[i][j])],
boggle, i, j, visited, str);
str = "";
}
}
}
}
// Driver program to test above function
int main()
{
// Let the given dictionary be following
char* dictionary[] = { "GEEKS", "FOR", "QUIZ", "GEE" };
// root Node of trie
TrieNode* root = getNode();
// insert all words of dictionary into trie
int n = sizeof(dictionary) / sizeof(dictionary[0]);
for (int i = 0; i < n; i++)
insert(root, dictionary[i]);
char boggle[M][N] = { { 'G', 'I', 'Z' },
{ 'U', 'E', 'K' },
{ 'Q', 'S', 'E' } };
findWords(boggle, root);
return 0;
} Java
// Java program for Boggle game
public class Boggle {
// Alphabet size
static final int SIZE = 26;
static final int M = 3;
static final int N = 3;
// trie Node
static class TrieNode {
TrieNode[] Child = new TrieNode[SIZE];
// isLeaf is true if the node represents
// end of a word
boolean leaf;
// constructor
public TrieNode()
{
leaf = false;
for (int i = 0; i < SIZE; i++)
Child[i] = null;
}
}
// If not present, inserts a key into the trie
// If the key is a prefix of trie node, just
// marks leaf node
static void insert(TrieNode root, String Key)
{
int n = Key.length();
TrieNode pChild = root;
for (int i = 0; i < n; i++) {
int index = Key.charAt(i) - 'A';
if (pChild.Child[index] == null)
pChild.Child[index] = new TrieNode();
pChild = pChild.Child[index];
}
// make last node as leaf node
pChild.leaf = true;
}
// function to check that current location
// (i and j) is in matrix range
static boolean isSafe(int i, int j, boolean visited[][])
{
return (i >= 0 && i < M && j >= 0
&& j < N && !visited[i][j]);
}
// A recursive function to print
// all words present on boggle
static void searchWord(TrieNode root, char boggle[][], int i,
int j, boolean visited[][], String str)
{
// if we found word in trie / dictionary
if (root.leaf == true)
System.out.println(str);
// If both I and j in range and we visited
// that element of matrix first time
if (isSafe(i, j, visited)) {
// make it visited
visited[i][j] = true;
// traverse all child of current root
for (int K = 0; K < SIZE; K++) {
if (root.Child[K] != null) {
// current character
char ch = (char)(K + 'A');
// Recursively search reaming character of word
// in trie for all 8 adjacent cells of
// boggle[i][j]
if (isSafe(i + 1, j + 1, visited)
&& boggle[i + 1][j + 1] == ch)
searchWord(root.Child[K], boggle,
i + 1, j + 1,
visited, str + ch);
if (isSafe(i, j + 1, visited)
&& boggle[i][j + 1] == ch)
searchWord(root.Child[K], boggle,
i, j + 1,
visited, str + ch);
if (isSafe(i - 1, j + 1, visited)
&& boggle[i - 1][j + 1] == ch)
searchWord(root.Child[K], boggle,
i - 1, j + 1,
visited, str + ch);
if (isSafe(i + 1, j, visited)
&& boggle[i + 1][j] == ch)
searchWord(root.Child[K], boggle,
i + 1, j,
visited, str + ch);
if (isSafe(i + 1, j - 1, visited)
&& boggle[i + 1][j - 1] == ch)
searchWord(root.Child[K], boggle,
i + 1, j - 1,
visited, str + ch);
if (isSafe(i, j - 1, visited)
&& boggle[i][j - 1] == ch)
searchWord(root.Child[K], boggle,
i, j - 1,
visited, str + ch);
if (isSafe(i - 1, j - 1, visited)
&& boggle[i - 1][j - 1] == ch)
searchWord(root.Child[K], boggle,
i - 1, j - 1,
visited, str + ch);
if (isSafe(i - 1, j, visited)
&& boggle[i - 1][j] == ch)
searchWord(root.Child[K], boggle,
i - 1, j,
visited, str + ch);
}
}
// make current element unvisited
visited[i][j] = false;
}
}
// Prints all words present in dictionary.
static void findWords(char boggle[][], TrieNode root)
{
// Mark all characters as not visited
boolean[][] visited = new boolean[M][N];
TrieNode pChild = root;
String str = "";
// traverse all matrix elements
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// we start searching for word in dictionary
// if we found a character which is child
// of Trie root
if (pChild.Child[(boggle[i][j]) - 'A'] != null) {
str = str + boggle[i][j];
searchWord(pChild.Child[(boggle[i][j]) - 'A'],
boggle, i, j, visited, str);
str = "";
}
}
}
}
// Driver program to test above function
public static void main(String args[])
{
// Let the given dictionary be following
String dictionary[] = { "GEEKS", "FOR", "QUIZ", "GEE" };
// root Node of trie
TrieNode root = new TrieNode();
// insert all words of dictionary into trie
int n = dictionary.length;
for (int i = 0; i < n; i++)
insert(root, dictionary[i]);
char boggle[][] = { { 'G', 'I', 'Z' },
{ 'U', 'E', 'K' },
{ 'Q', 'S', 'E' } };
findWords(boggle, root);
}
}
// This code is contributed by Sumit GhoshC#
// C# program for Boggle game
using System;
public class Boggle {
// Alphabet size
static readonly int SIZE = 26;
static readonly int M = 3;
static readonly int N = 3;
// trie Node
public class TrieNode {
public TrieNode[] Child = new TrieNode[SIZE];
// isLeaf is true if the node represents
// end of a word
public bool leaf;
// constructor
public TrieNode()
{
leaf = false;
for (int i = 0; i < SIZE; i++)
Child[i] = null;
}
}
// If not present, inserts a key into the trie
// If the key is a prefix of trie node, just
// marks leaf node
static void insert(TrieNode root, String Key)
{
int n = Key.Length;
TrieNode pChild = root;
for (int i = 0; i < n; i++) {
int index = Key[i] - 'A';
if (pChild.Child[index] == null)
pChild.Child[index] = new TrieNode();
pChild = pChild.Child[index];
}
// make last node as leaf node
pChild.leaf = true;
}
// function to check that current location
// (i and j) is in matrix range
static bool isSafe(int i, int j, bool[, ] visited)
{
return (i >= 0 && i < M && j >= 0 && j < N && !visited[i, j]);
}
// A recursive function to print all words present on boggle
static void searchWord(TrieNode root, char[, ] boggle, int i,
int j, bool[, ] visited, String str)
{
// if we found word in trie / dictionary
if (root.leaf == true)
Console.WriteLine(str);
// If both I and j in range and we visited
// that element of matrix first time
if (isSafe(i, j, visited)) {
// make it visited
visited[i, j] = true;
// traverse all child of current root
for (int K = 0; K < SIZE; K++) {
if (root.Child[K] != null) {
// current character
char ch = (char)(K + 'A');
// Recursively search reaming character of word
// in trie for all 8 adjacent cells of
// boggle[i, j]
if (isSafe(i + 1, j + 1, visited) && boggle[i + 1, j + 1] == ch)
searchWord(root.Child[K], boggle, i + 1, j + 1,
visited, str + ch);
if (isSafe(i, j + 1, visited) && boggle[i, j + 1] == ch)
searchWord(root.Child[K], boggle, i, j + 1,
visited, str + ch);
if (isSafe(i - 1, j + 1, visited) && boggle[i - 1, j + 1] == ch)
searchWord(root.Child[K], boggle, i - 1, j + 1,
visited, str + ch);
if (isSafe(i + 1, j, visited) && boggle[i + 1, j] == ch)
searchWord(root.Child[K], boggle, i + 1, j,
visited, str + ch);
if (isSafe(i + 1, j - 1, visited) && boggle[i + 1, j - 1] == ch)
searchWord(root.Child[K], boggle, i + 1, j - 1,
visited, str + ch);
if (isSafe(i, j - 1, visited) && boggle[i, j - 1] == ch)
searchWord(root.Child[K], boggle, i, j - 1,
visited, str + ch);
if (isSafe(i - 1, j - 1, visited) && boggle[i - 1, j - 1] == ch)
searchWord(root.Child[K], boggle, i - 1, j - 1,
visited, str + ch);
if (isSafe(i - 1, j, visited) && boggle[i - 1, j] == ch)
searchWord(root.Child[K], boggle, i - 1, j,
visited, str + ch);
}
}
// make current element unvisited
visited[i, j] = false;
}
}
// Prints all words present in dictionary.
static void findWords(char[, ] boggle, TrieNode root)
{
// Mark all characters as not visited
bool[, ] visited = new bool[M, N];
TrieNode pChild = root;
String str = "";
// traverse all matrix elements
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// we start searching for word in dictionary
// if we found a character which is child
// of Trie root
if (pChild.Child[(boggle[i, j]) - 'A'] != null) {
str = str + boggle[i, j];
searchWord(pChild.Child[(boggle[i, j]) - 'A'],
boggle, i, j, visited, str);
str = "";
}
}
}
}
// Driver program to test above function
public static void Main(String[] args)
{
// Let the given dictionary be following
String[] dictionary = { "GEEKS", "FOR", "QUIZ", "GEE" };
// root Node of trie
TrieNode root = new TrieNode();
// insert all words of dictionary into trie
int n = dictionary.Length;
for (int i = 0; i < n; i++)
insert(root, dictionary[i]);
char[, ] boggle = { { 'G', 'I', 'Z' },
{ 'U', 'E', 'K' },
{ 'Q', 'S', 'E' } };
findWords(boggle, root);
}
}
// This code has been contributed by 29AjayKumar输出：
GEE, GEEKS, QUIZ
复杂度分析：
- 时间复杂度： O(4 ^(N ^ 2))。
即使应用了trie，时间复杂度也保持不变。对于每个单元，有4个方向，并且有N ^ 2个单元。因此，时间复杂度为O(4 ^(N ^ 2))。 - 辅助空间： O(N ^ 2)。
递归的最大长度可以为N ^ 2，其中N是矩阵的边。因此，空间复杂度为O(N ^ 2)。
如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。