Word search

来源：互联网发布：数据库的概念模型设计编辑：程序博客网时间：2024/06/06 15:39

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.

犯错的地方：

1. 在主函数里也要mark visit过的为".“

2. 各种下标总是copy后忘记改

public class Solution {    public boolean exist(char[][] board, String word) {        if (word.isEmpty()) {            return false;        }                int rowN = board.length;        int colN = board[0].length;        for (int i = 0; i < rowN; i++) {            for (int j = 0; j < colN; j++) {                if (board[i][j] == word.charAt(0)) {                    board[i][j] = '.';  <span style="color:#ff0000;">//这个地方忘了标记。只在辅助函数里记得标记了！</span>                    if (helper(board, word.substring(1), i, j)) {                        return true;                    } else {                        board[i][j] = word.charAt(0);                    }                }            }        }        return false;    }        boolean helper(char[][] board, String word, int row, int col) {        if (word.isEmpty()) {            return true;        }                char curChar = word.charAt(0);        //up        if (row > 0 && board[row-1][col] == curChar) {            board[row-1][col] = '.';            if (helper(board, word.substring(1), row-1, col)) {                return true;            } else {                board[row-1][col] = curChar;            }        }        //down        if (row < board.length-1 && board[row+1][col] == curChar) {            board[row+1][col] = '.';            if (helper(board, word.substring(1), row+1, col)) {                return true;            } else {                board[row+1][col] = curChar;            }        }                //left        if (col > 0 && board[row][col-1] == curChar) {            board[row][col-1] = '.';            if (helper(board, word.substring(1), row, col-1)) {                return true;            } else {                board[row][col-1] = curChar;            }        }        //right        if (col < board[0].length-1 && board[row][col+1] == curChar) {            board[row][col+1] = '.';            if (helper(board, word.substring(1), row, col+1)) {                return true;            } else {                board[row][col+1] = curChar;            }        }        return false;    }}

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