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。
例如,在下面的 3×4 的矩阵中包含单词 "ABCCED"(单词中的字母已标出)。
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","d"]], word = "abcd"
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
注意:本题与主站 79 题相同:https://leetcode-cn.com/problems/word-search/
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/ju-zhen-zhong-de-lu-jing-lcof
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一般这种题都能用回溯做。
个人感觉回溯最重要的一个是终止判断,另一个是状态恢复。
class Solution { public boolean exist(char[][] board, String word) { if (board == null || board.length == 0 || board[0].length == 0) { return false; } char[] arr = word.toCharArray(); boolean[][] judge = 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 (board[i][j] == arr[0]) { if (dfs(arr, 0, board, i, j, judge)) { return true; } } } } return false; } private boolean dfs(char[] arr, int p, char[][] board, int i, int j, boolean[][] judge) { if (p == arr.length) { return true; } if (i < 0 || i == board.length) { return false; } if (j < 0 || j == board[0].length) { return false; } if (judge[i][j]) { return false; } if (board[i][j] != arr[p]) { return false; } judge[i][j] = true; boolean ans = dfs(arr, p + 1, board, i + 1, j, judge) || dfs(arr, p + 1, board, i - 1, j, judge) || dfs(arr, p + 1, board, i, j + 1, judge) || dfs(arr, p + 1, board, i, j - 1, judge); judge[i][j] = false; return ans; } }
标签:arr,12,return,Offer,---,length,board,judge,false From: https://www.cnblogs.com/allWu/p/17279988.html