lintcode :单词搜索

题目

单词搜索

给出一个二维的字母板和一个单词，寻找字母板网格中是否存在这个单词。

单词可以由按顺序的相邻单元的字母组成，其中相邻单元指的是水平或者垂直方向相邻。每个单元中的字母最多只能使用一次。

样例

给出board =

[

"ABCE",

"SFCS",

"ADEE"

]

word = "ABCCED"， ->返回 true,

word = "SEE"，-> 返回 true,

word = "ABCB"， -> 返回 false.

解题

直接深度搜索

public class Solution {
    public boolean exist(char[][] board, String word) {
        return method(board,word);
    }
    public boolean method(char[][] board,String word){
        int row = board.length;
        int col = board[0].length;
        boolean[][] visited = new boolean[row][col];
        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++)
                if(dfs(board,visited,i,j,0,word))
                    return true;
        return false;
    }
    public boolean dfs(char[][] board,boolean[][] visited,int row,int col,int index,String word){
        if(word.length() == index){
            return true;
        }
        if(row<0 || col<0||row>=board.length || col>=board[0].length) return false;
        char ch = word.charAt(index);
        if(!visited[row][col] && ch == board[row][col]){
            visited[row][col] = true;
            boolean res = dfs(board,visited,row-1,col,index+1,word)|| dfs(board,visited,row+1,col,index+1,word)
                   ||dfs(board,visited,row,col-1,index+1,word)|| dfs(board,visited,row,col+1,index+1,word);
            visited[row][col] = false;
            return res;
        }
        return false;
    }
}

Java Code

修改原始数组，降低空间复杂度

public class Solution {
    /**
     * @param board: A list of lists of character
     * @param word: A string
     * @return: A boolean
     */
    public boolean exist(char[][] board, String word) {
        // write your code here
        return method(board,word);
    }
    public boolean method(char[][] board,String word){
        int row = board.length;
        int col = board[0].length;

        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++){
                if(dfs(board,i,j,0,word))
                    return true;

            }
        return false;
    }
    public boolean dfs(char[][] board,int row,int col,int index,String word){
        if(word.length() == index){
            return true;
        }
        if(row<0 || col<0||row>=board.length || col>=board[0].length) return false;
        char ch = word.charAt(index);
        if(board[row][col]!='+' && ch == board[row][col]){
            char c = board[row][col];
            board[row][col]='+';
            boolean res = dfs(board,row-1,col,index+1,word)|| dfs(board,row+1,col,index+1,word)
                   ||dfs(board,row,col-1,index+1,word)|| dfs(board,row,col+1,index+1,word);
            board[row][col] = c;
            return res;
        }
        return false;
    }
}

Java Code

Python

class Solution(object):
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        for i in xrange(len(board)):
            for j in xrange(len(board[0])):
                if self.dfs(board,i,j,word,0):
                    return True
        return False 

    def dfs(self,board,row,col,word,index):
        if index == len(word):
            return True
        if row<0 or col<0 or row>=len(board) or col>=len(board[0]):
            return False
        ch = word[index]
        res = False
        if(board[row][col]!='+' and board[row][col] ==ch):
            c = board[row][col]
            board[row][col] = '+'
            res = self.dfs(board,row-1,col,word,index+1) or \
                  self.dfs(board,row,col-1,word,index+1) or \
                  self.dfs(board,row+1,col,word,index+1) or \
                  self.dfs(board,row,col+1,word,index+1)
            if res:
                return res
            else:
                board[row][col] = c
        return False