[LeetCode] 79. Word Search 单词搜索

2021年09月15日 阅读数:1
这篇文章主要向大家介绍[LeetCode] 79. Word Search 单词搜索,主要内容包括基础应用、实用技巧、原理机制等方面,希望对大家有所帮助。

Given a 2D board and a word, find if the word exists in the grid.html

The word can 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.java

For example,
Given board =python

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.数组

给定一个2维的字母board,判断 是否有一个网格路径组成给定的单词。post

解法:DFS, 典型的深度优先遍历,对每一点的每一条路径进行深度遍历,遍历过程当中一旦出现:url

1.数组越界。2.该点已访问过。3.该点的字符和word对应的index字符不匹配。spa

就要对该路径进行剪枝:设计

Java:code

public boolean exist(char[][] board, String word) {
    int m = board.length;
    int n = board[0].length;
 
    boolean result = false;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
           if(dfs(board,word,i,j,0)){
               result = true;
           }
        }
    }
 
    return result;
}
 
public boolean dfs(char[][] board, String word, int i, int j, int k){
    int m = board.length;
    int n = board[0].length;
 
    if(i<0 || j<0 || i>=m || j>=n){
        return false;
    }
 
    if(board[i][j] == word.charAt(k)){
        char temp = board[i][j];
        board[i][j]='#';
        if(k==word.length()-1){
            return true;
        }else if(dfs(board, word, i-1, j, k+1)
        ||dfs(board, word, i+1, j, k+1)
        ||dfs(board, word, i, j-1, k+1)
        ||dfs(board, word, i, j+1, k+1)){
            return true;
        }
        board[i][j]=temp;
    }
 
    return false;
} 

Java:htm

class Solution {
    int[] dh = {0, 1, 0, -1};  
    int[] dw = {1, 0, -1, 0};
  
    public boolean exist(char[][] board, String word) {  
        boolean[][] isVisited = new boolean[board.length][board[0].length];  
        for (int i = 0; i < board.length; i++)  
            for (int j = 0; j < board[0].length; j++)  
                if (isThisWay(board, word, i, j, 0, isVisited)) return true;  
        return false;  
    }  
  
    public boolean isThisWay(char[][] board, String word, int row, int column, int index, boolean[][] isVisited) {  
        if (row < 0 || row >= board.length || column < 0 || column >= board[0].length  
            || isVisited[row][column] || board[row][column] != word.charAt(index))  
                return false;  //剪枝  
        if (++index == word.length()) return true;  //word全部字符均匹配上  
        isVisited[row][column] = true;  
        for (int i = 0; i < 4; i++)  
            if (isThisWay(board, word, row + dh[i], column + dw[i], index, isVisited))  
                return true;  //以board[row][column]为起点找到匹配上word路径  
        isVisited[row][column] = false;  //遍历事后,将该点还原为未访问过  
        return false;  
    } 
}  

Python:

class Solution:
    # @param board, a list of lists of 1 length string
    # @param word, a string
    # @return a boolean
    def exist(self, board, word):
        visited = [[False for j in xrange(len(board[0]))] for i in xrange(len(board))]
        
        for i in xrange(len(board)):
            for j in xrange(len(board[0])):
                if self.existRecu(board, word, 0, i, j, visited):
                    return True
        
        return False
    
    def existRecu(self, board, word, cur, i, j, visited):
        if cur == len(word):
            return True
        
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or visited[i][j] or board[i][j] != word[cur]:
            return False
        
        visited[i][j] = True
        result = self.existRecu(board, word, cur + 1, i + 1, j, visited) or\
                 self.existRecu(board, word, cur + 1, i - 1, j, visited) or\
                 self.existRecu(board, word, cur + 1, i, j + 1, visited) or\
                 self.existRecu(board, word, cur + 1, i, j - 1, visited)         
        visited[i][j] = False
        
        return result

C++:

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        if (word.empty()) return true;
        if (board.empty() || board[0].empty()) return false;
        vector<vector<bool> > visited(board.size(), vector<bool>(board[0].size(), false));
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if (search(board, word, 0, i, j, visited)) return true;
            }
        }
        return false;
    }
    bool search(vector<vector<char> > &board, string word, int idx, int i, int j, vector<vector<bool> > &visited) {
        if (idx == word.size()) return true;
        if (i < 0 || j < 0 || i >= board.size() || j >= board[0].size() || visited[i][j] || board[i][j] != word[idx]) return false;
        visited[i][j] = true;
        bool res = search(board, word, idx + 1, i - 1, j, visited) 
                 || search(board, word, idx + 1, i + 1, j, visited)
                 || search(board, word, idx + 1, i, j - 1, visited)
                 || search(board, word, idx + 1, i, j + 1, visited);
        visited[i][j] = false;
        return res;
    }
}; 

 

相似题目:

[LeetCode] 212. Word Search II 词语搜索 II

[LeetCode] 348. Design Tic-Tac-Toe 设计井字棋游戏

 

 

All LeetCode Questions List 题目汇总