• [LeetCode] Word Search


    Question:

    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 =

    [
      ["ABCE"],
      ["SFCS"],
      ["ADEE"]
    ]
    

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

    1、题型分类:

    2、思路:

      该题类似于迷宫,递归回溯。要实现从某一步开始往下找,发现找不到了,能够回到当初的那一步,那么就要将递归的地方放在搜索的过程中,要判断能否搜索到结果,就要将递归放在if语句中,搜索失败后能从这一步再次搜索。要实现不搜索找过的路,需要一个辅助数组。

    3、时间复杂度:

    4、代码:

        public boolean exist(char[][] board, String word) {
            //an array help to determine the path has searched
            if(board==null || word==null || word.length()==0) return false;
            //if(board.length==1) return new String(board[0]).indexOf(word)>=0;
            boolean [][] check=new boolean [board.length][board[0].length];
            for(int i=0;i<board.length;i++)
            {
                for(int j=0;j<board[0].length;j++)
                {
                    check[i][j]=false;
                }
            }
            for(int i=0;i<board.length;i++)
            {
                for(int j=0;j<board[0].length;j++)
                {
                    if(board[i][j]==word.charAt(0) && check[i][j]==false)
                    {
                        check[i][j]=true;
                        if(word.length()==1 || search(board, word.substring(1), i, j, check))
                            return true;
                        check[i][j]=false;
                    }
                }
            }
            return false;
        }
        public boolean search(char[][] board, String word,int y,int x,boolean[][] check)
        {
            int[][] direction={{1,0},{-1,0},{0,1},{0,-1}};
            for(int k=0;k<direction.length;k++)
            {
                int xx=x+direction[k][0];
                int yy=y+direction[k][1];
                if(xx>=0 && xx<board[0].length
                    && yy>=0 && yy<board.length
                    && board[yy][xx]==word.charAt(0) 
                    && check[yy][xx]==false)
                {
                    check[yy][xx]=true;
                    if(word.length()==1 || search(board, word.substring(1), yy, xx, check))
                        return true;
                    check[yy][xx]=false;
                }
            }
            return false;
        }

    5、优化:

    6、扩展:

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  • 原文地址:https://www.cnblogs.com/maydow/p/4646307.html
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