刷题61 单词搜索

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：

输入： board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出： true

示例 2：

输入： board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出： true

示例 3：

输入： board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出： false

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

进阶： 你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？

题解：

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

    visited[i][j]=true;
    int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    boolean result = false;
    for (int[] dir : directions) {
        int newi = i + dir[0], newj = j + dir[1];
        if (newi >= 0 && newi < board.length && newj >= 0 && newj < board[0].length) {
            if (!visited[newi][newj]) {
                boolean flag = dfs(board,word, idx + 1,newi, newj, visited );
                if (flag) {
                    result = true;
                    break;
                }
            }
        }
    }
    visited[i][j] = false;
    return result;

}

}