力扣刷题:单词搜索(java实现)

题目: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:


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

示例 2:


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

示例 3:


image.png
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?

相关标签:数组回溯矩阵

解析:这题大体思路就是先找到单词的起点位置,然后根据上下左右四个方向找符合条件的字符,同时考虑到不能重复使用同一个位置的字符这个条件,可以把访问过的路径的字符置为*号,遍历过后替换回来。具体代码如下所示:

public boolean exist(char[][] board, String word) {
        char[] arr = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if(dfs(board,arr,i,j,0)){
                    return true;
                }
            }
        }
        return false;
    }

    public boolean dfs(char[][] board,char[] arr,int i,int j,int index){
        //判断边界条件(中断条件)
        if(i<0 || i>=board.length || j<0 || j>=board[0].length || board[i][j]!=arr[index]){
            return false;
        }
        //如果遍历完成(结束条件)
        if(index == arr.length-1){
            return true;
        }
        //保存当前表格里的字符
        char temp = board[i][j];
        //将走过的路径设置成*,防止重复
        board[i][j] = '*';
        //继续判断上下左右四个方向
        boolean top  = dfs(board,arr,i-1,j,index+1);
        boolean bottom  = dfs(board,arr,i+1,j,index+1);
        boolean right  = dfs(board,arr,i,j+1,index+1);
        boolean left  = dfs(board,arr,i,j-1,index+1);
        //将之前替换的字符还原
        board[i][j] = temp;
        return top || bottom || right || left;
    }

每天进步一点点~

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