【重点】【回溯】【DFS】79.单词搜索

题目
注意：此题跟岛屿的数量对比来看，增加了回溯的过程，岛屿题并无回溯。

法1：DFS

必须掌握方法！

class Solution {
   
    public boolean exist(char[][] board, String word) {
   
        char[] charArr = word.toCharArray();
        int m = board.length, n = board[0].length;
        for (int i = 0; i < m; ++i) {
   
            for (int j = 0; j < n; ++j) {
   
                int[][] used = new int[m][n];
                if (dfs(board, charArr, used, i, j, 0)) {
   
                    return true;
                }
            }
        }

        return false;
    }

    public boolean dfs(char[][] board, char[] charArr, int[][] used, int i, int j, int curIndex) {
   
        if (i < 0 || i >= board.length 
                || j < 0 || j >= board[0].length 
                || used[i][j] == 1
                || curIndex >= charArr.length 
                || board[i][j] != charArr[curIndex]) {
   
            return false;
        }
        if (curIndex == charArr.length - 1 && board[i][j] == charArr[curIndex]) {
   
            return true;
        }
        used[i][j] = 1;

        boolean res = dfs(board, charArr, used, i - 1, j, curIndex + 1) 
                        || dfs(board, charArr, used, i + 1, j, curIndex + 1) 
                        || dfs(board, charArr, used, i, j - 1, curIndex + 1) 
                        || dfs(board, charArr, used, i, j + 1, curIndex + 1);
        if (res) {
   
            return true;
        } else {
   
            used[i][j] = 0;
            return false;
        }
    }
}