英文搜索网站_DFS搜索

编程汇总 • 2025-01-19 20:01 • 阅读 9

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words，找出所有同时在二维网格和字典中出现的单词。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

示例 1：





输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]

输出：["eat","oath"]

示例 2：





输入：board = [["a","b"],["c","d"]], words = ["abcb"]

输出：[]

提示：

m == board.length
n == board[i].length
1 <= m, n <= 12
board[i][j] 是一个小写英文字母
1 <= words.length <= 3 * 104
1 <= words[i].length <= 10
words[i] 由小写英文字母组成
words 中的所有字符串互不相同
题解
Trie+暴力搜索

class Solution { 

   

public:

    static const int N = 3e4 + 10;

    int trie[N][26],cnt[N],idx;

    void insert(string &t){ 

   

        int p = 0;

        for(auto &c : t){ 

   

            int u = c - 'a';

            if(trie[p][u] == 0)trie[p][u] = ++idx;

            p = trie[p][u];

            if(cnt[p] != 1)cnt[p] = -1;

        }

        cnt[p] = 1;

    }

    vectorres;

    string t = "";

    int n,m;

    int vis[12][12],dx[4] = { 

   0,1,0,-1},dy[4] = { 

   -1,0,1,0};

    int flag[N] = { 

   0};

    setss;

    void dfs(int x,int y,vector>& board,int p){ 

   

        if(cnt[p] == 1){ 

   

            if(flag[p] == 0)res.push_back(t),flag[p] = 1;

        }else if(cnt[p] == 0){ 

   

            return;

        }

        if(x < 0 || x >= n || y < 0 || y >= m)return;

        if(vis[x][y])return;

        vis[x][y] = true;

        t.append(1,board[x][y]);

        for(int k = 0;k < 4;k ++){ 

   

            int a = x + dx[k],b = y + dy[k];

            if(trie[p][board[x][y] - 'a'] != 0)

            dfs(a,b,board,trie[p][board[x][y] - 'a']);

        }

        t.erase(t.size() - 1,1);

        vis[x][y] = false;

    }

    vector findWords(vector>& board, vector& words) { 

   

        for(auto &word : words)insert(word);

        cnt[0] = -1;

        n = board.size(),m = board[0].size();

        for(int i = 0;i < n;i ++){ 

   

            for(int j = 0;j < m;j ++){ 

   

                dfs(i,j,board,0);

            }

        }

        return res;

    }

};

英文搜索网站_DFS搜索

相关推荐