leetcode79. Word Search

79. Word Search

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.

解法

逐个遍历矩阵中的字符，以该字符开头，进行dfs查找，围绕该元素四周进行判断，是否存在下一个元素，不能回头。其中不能回头，用一个矩阵对每一个元素做标记，已经访问过为true，没有访问过为false，当某种情况走不通时，所有的元素的标记都还原为false。最上层再遍历下一个元素。

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