[LeetCode] 79. Word Search 單詞搜索

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

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.

For example,
Given board =

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

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

給定一個2維的字母board，判斷 是否有一個網格路徑組成給定的單詞。

解法：DFS， 典型的深度優先遍歷，對每一點的每一條路徑進行深度遍歷，遍歷過程當中一旦出現：

1.數組越界。2.該點已訪問過。3.該點的字符和word對應的index字符不匹配。

就要對該路徑進行剪枝：

Java:

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：

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]爲起點找到匹配上word路徑  
        isVisited[row][column] = false;  //遍歷事後，將該點還原爲未訪問過  
        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;
    }
};　

 

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