https://leetcode.com/problems/word-search/

class Solution {
public:
struct Trie{
Trie *next[];
bool isWord;
Trie() {
this->isWord = false;
for(auto &c: next) c = NULL;
}
};
void insert(Trie *root, string word) {
Trie *p = root;
for(auto &c: word) {
int idx = c - 'A';
if(!p->next[idx]) p->next[idx] = new Trie();
p = p->next[idx];
}
p->isWord = true;
}
bool isPrefix(Trie *root, string word) {
Trie *p = root;
for(auto &c: word) {
int idx = c - 'A';
if(!p->next[idx]) return false;
p = p->next[idx];
}
return true;
}
bool isWord(Trie *root, string word) {
Trie *p = root;
for(auto &c: word) {
int idx = c - 'A';
if(!p->next[idx]) return false;
p = p->next[idx];
}
return p->isWord;
}
bool check(vector<vector<char>> &board, int x, int y) {
int m = board.size(), n = board[].size();
if(x< || y< || x>=m || y>=n) return false;
return true;
} bool dfs(vector<vector<char>> &board, vector<vector<bool>> &vis, Trie *root, string s, int x, int y) {
if(isWord(root, s))
return true;
} int dir[][] = {{,}, {,}, {-,}, {,-}};
for(int i=;i<;++i) {
int nx = x + dir[i][], ny = y + dir[i][];
if(check(board, nx, ny) && !vis[nx][ny]) {
if(isPrefix(root, s + board[nx][ny])) {
vis[nx][ny] = true;
if(dfs(board, vis, root, s + board[nx][ny], nx, ny)) return true;
vis[nx][ny] = false;
}
}
}
return false;
}
bool exist(vector<vector<char>>& board, string word) {
if (board.empty() || board[].empty() || word.empty()) return false; string s = "";
int m = board.size(), n = board[].size();
vector<vector<bool>> vis(m, vector<bool>(n, false)); Trie *root = new Trie();
insert(root, word); for(int i=;i<m;++i) {
for(int j=;j<n;++j) {
//if(isWord(root, s + board[i][j])) return true;
if(isPrefix(root, s)) {
vis[i][j] = true;
if(dfs(board, vis, root, s + board[i][j], i, j)) return true;
vis[i][j] = false;
}
}
} return false;
}
};

https://leetcode.com/problems/word-search-ii/

struct TriNode {
TriNode *ch[];
bool isWord;
TriNode() : isWord(false) {
for (auto &a : ch) a = NULL;
}
};
class Solution {
public:
void insert(TriNode *root, string word) {
TriNode *p = root;
for (auto &a : word) {
int i = a - 'a';
if (p->ch[i] == NULL) p->ch[i] = new TriNode();
p = p->ch[i];
}
p->isWord = true;
}
bool isPrefix(TriNode *root, string word) {
TriNode *p = root;
for (auto &a : word) {
int i = a - 'a';
if (p->ch[i] == NULL) return false;
p = p->ch[i];
}
return true;
}
bool isWord(TriNode *root, string word) {
TriNode *p = root;
for (auto &a : word) {
int i = a - 'a';
if (p->ch[i] == NULL) return false;
p = p->ch[i];
}
return p->isWord;
}
bool isValid(vector<vector<char>> &board, int x, int y) {
int m = board.size(), n = board[].size();
if (x < || x >= m || y < || y >= n) return false;
return true;
}
void dfs(vector<vector<char>> board, vector<vector<int>> visit, set<string> &st, string s, TriNode *root, int x, int y) {
if(!isPrefix(root, s)) return;
if(isWord(root, s)) {
st.insert(s);
} int dx[] = {, -, , };
int dy[] = {, , , -};
int xx, yy;
for (int i = ; i < ; ++i) {
xx = x + dx[i]; yy = y + dy[i];
if (isValid(board, xx, yy) && !visit[xx][yy]) {
visit[xx][yy] = ;
dfs(board, visit, st, s + board[xx][yy], root, xx, yy);
visit[xx][yy] = ;
}
}
} vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
vector<string> res; res.clear();
if (board.empty() || board[].empty() || words.empty()) return res; TriNode *root = new TriNode();
for(auto &word : words) insert(root, word); int m = board.size(), n = board[].size();
vector<vector<int>> visit(m, vector<int>(n, )); string s="";
set<string> st; st.clear();
for (int i = ; i < m; ++i) {
for (int j = ; j < n; ++j) {
visit[i][j] = ;
dfs(board, visit, st, s + board[i][j], root, i, j);
visit[i][j] = ;
}
} for (auto &word : st) res.push_back(word);
return res;
}
};

leetcode@ [79/140] Trie树应用 Word Search / Word Search II的更多相关文章

  1. 字典树(查找树) leetcode 208. Implement Trie (Prefix Tree) 、211. Add and Search Word - Data structure design

    字典树(查找树) 26个分支作用:检测字符串是否在这个字典里面插入.查找 字典树与哈希表的对比:时间复杂度:以字符来看:O(N).O(N) 以字符串来看:O(1).O(1)空间复杂度:字典树远远小于哈 ...

  2. leetcode 211. Add and Search Word - Data structure design Trie树

    题目链接 写一个数据结构, 支持两种操作. 加入一个字符串, 查找一个字符串是否存在.查找的时候, '.'可以代表任意一个字符. 显然是Trie树, 添加就是正常的添加, 查找的时候只要dfs查找就可 ...

  3. leetcode 79. Word Search 、212. Word Search II

    https://www.cnblogs.com/grandyang/p/4332313.html 在一个矩阵中能不能找到string的一条路径 这个题使用的是dfs.但这个题与number of is ...

  4. LeetCode 79 Word Search(单词查找)

    题目链接:https://leetcode.com/problems/word-search/#/description 给出一个二维字符表,并给出一个String类型的单词,查找该单词是否出现在该二 ...

  5. [leetcode trie]211. Add and Search Word - Data structure design

    Design a data structure that supports the following two operations: void addWord(word) bool search(w ...

  6. [LeetCode] 79. Word Search 单词搜索

    Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from l ...

  7. [leetcode]79.Search Word 回溯法

    /** * Given a 2D board and a word, find if the word exists in the grid. The word can be constructed ...

  8. [LeetCode] Add and Search Word - Data structure design 添加和查找单词-数据结构设计

    Design a data structure that supports the following two operations: void addWord(word) bool search(w ...

  9. leetcode@ [211] Add and Search Word - Data structure design

    https://leetcode.com/problems/add-and-search-word-data-structure-design/ 本题是在Trie树进行dfs+backtracking ...

随机推荐

  1. uva 10940

    数学 打了个表 找一下规律.... #include <cstdio> int a[30]; void init() { a[1]=2;a[2]=4;a[3]=8;a[4]=16;a[5] ...

  2. Creating a new Signiant Transfer Engine because the previous transfer had to be canceled.

    From: http://stackoverflow.com/questions/10548196/application-loader-new-weird-warning-about-signian ...

  3. Python利用ConfigParser读取配置文件

    http://www.2cto.com/kf/201108/100384.html #!/usr/bin/python # -*- coding:utf-8 -*- import ConfigPars ...

  4. overrides final method getUnknownFields.()Lcom/google/protobuf/UnknownFieldSet 错误解决

    使用java代码连接hbase服务器报错:  java.lang.VerifyError: class org.apache.hadoop.hdfs.protocol.proto.ClientName ...

  5. 李洪强iOS开发之【零基础学习iOS开发】【02-C语言】08-基本运算

    计算机的基本能力就是计算,所以一门程序设计语言的计算能力是非常重要的.C语言之所以无所不能,是因为它不仅有丰富的数据类型,还有强大的计算能力.C语言一共有34种运算符,包括了常见的加减乘除运算.这讲就 ...

  6. UNIX内核的文件数据结构 -- v 节点与 i 节点

    龙泉居士:http://hi.baidu.com/zeyu203/item/cc89cfc0f36bfecc994aa07c 内核使用三种数据结构表示打开的文件(如图),他们之间的关系决定了在文件共享 ...

  7. DSPLIB for C6455+CCSv3.3

    问题描述: Hello everybody, I was looking for DSPLIB libraries optimized for C6455 processors. I found th ...

  8. API HOOK技术

    API HOOK技术是一种用于改变API执行结果的技术,Microsoft 自身也在Windows操作系统里面使用了这个技术,如Windows兼容模式等. API HOOK 技术并不是计算机病毒专有技 ...

  9. sed找到重复的行

    sed之仅打印相邻重复的行 cat file  aaa bbb bbb ccc ddd eee eee fff   只显示重复的行: bbb bbb eee eee   sed -n ':a;N;/\ ...

  10. node安装插件方法

    node安装插件方法有几种,这里列出常用的两种方法: 方法1: 进入要安装插件的目录,直接用 npm 软件安装包安装,如(安装express): cd /project npm install -g ...