Word Search II
Desicription
Given a 2D board and a list of words from the dictionary, find all words in the board.
Each word must be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
Example:
代码语言：javascript复制Input:
words = ["oath","pea","eat","rain"] and board =
[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]
Output: ["eat","oath"]Note:
You may assume that all inputs are consist of lowercase letters a-z.
Solution
代码语言：javascript复制class Solution {
private:
vector<vector<int>> dir{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
struct TrieNode{
char value;
vector<TrieNode*> children;
explicit TrieNode(char c):value(c){};
};
TrieNode* root = new TrieNode('#');
vector<string> result;
void dfs(vector<vector<char>>& board, int x, int y, string currentString, TrieNode* currentNode) {
for(auto& childrenNode : currentNode->children) {
if(board[x][y] == childrenNode->value) {
currentString = board[x][y];
for(auto& nextNode : childrenNode->children) {
if(nextNode->value == '#') {
result.push_back(currentString);
childrenNode->children.erase(find(childrenNode->children.begin(), childrenNode->children.end(), nextNode));
}
}
board[x][y] ^= 255;
for(int i = 0; i < 4; i ) {
int dx = x dir[i][0];
int dy = y dir[i][1];
if(dx < 0 || dx >= board.size() || dy < 0 || dy >= board[0].size()) {
continue;
}
dfs(board, dx, dy, currentString, childrenNode);
}
board[x][y] ^= 255;
}
}
}
public:
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
set<string> wordsSet;
for(const auto ¤tWord : words) {
wordsSet.insert(currentWord);
}
for(auto currentWord : wordsSet) {
if(currentWord.empty()) {
result.emplace_back("");
continue;
}
currentWord = "#";
int index = 0;
auto pre = root;
while(currentWord[index]) {
bool found = false;
for(auto children : pre->children) {
if(children->value == currentWord[index]) {
found = true;
pre = children;
break;
}
}
if(!found) {
auto nextNode = new TrieNode(currentWord[index]);
pre->children.push_back(nextNode);
pre = nextNode;
}
index ;
}
}
if(board.empty() || board[0].empty()) {
return result;
}
for(int i = 0; i < board.size(); i ) {
for(int j = 0; j < board[0].size(); j ) {
dfs(board, i, j, "", root);
}
}
return result;
}
};
