LeetCode刷题【79】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？

标准回溯题

1. 挨个遍历找首字母匹配
2. 从匹配的首字母开始往四个方向遍历匹配下一个字母，并记录已访问过的点坐标
3. 遍历的过程中如果有字母不匹配返回false
4. 遍历的过程中判断越界情况和已访问过的情况
5. 直到匹配到最后一个字符完全相同 返回false
6. 退出递归记得改掉已访问过的坐标为false

class Solution {

    char[][] board;
    int[][] target = {{1,0},{0,1},{-1,0},{0,-1}};
    String word;

    public boolean exist(char[][] board, String word) {
        this.board = board;
        this.word = word;
        boolean[][] visited = new boolean[board.length][board[0].length];
        for (int row = 0; row < board.length; row++) {
            for (int col = 0; col < board[0].length; col++) {
                visited[row][col] = true;
                if (board[row][col] == word.charAt(0) && dfs(row,col, visited, 0)){
                    return true;
                }
                visited[row][col] = false;
            }
        }
        return false;
    }



    private boolean dfs(int row, int col, boolean[][] visited, int idx){
        if (idx== word.length()-1 && word.charAt(idx)==board[row][col]){
            return true;
        }

        if (word.charAt(idx) != board[row][col]){
            return false;
        }
        for (int[] ints : target) {
            int nextRow = row+ints[0];
            int nextCol = col+ints[1];
            if (nextRow>= board.length || nextRow < 0 || nextCol >= board[0].length || nextCol < 0 || visited[nextRow][nextCol]){
                continue;
            }
            visited[nextRow][nextCol] = true;
            if (dfs(nextRow,nextCol, visited,idx+1)){
                return true;
            }
            visited[nextRow][nextCol] = false;
        }
        return false;
    }
}