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  • 单词搜索

    单词搜索

    题目:
    给定一个二维网格和一个单词,找出该单词是否存在于网格中。

    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:

    board =
    [
    ['A','B','C','E'],
    ['S','F','C','S'],
    ['A','D','E','E']
    ]

    给定 word = "ABCCED", 返回 true
    给定 word = "SEE", 返回 true
    给定 word = "ABCB", 返回 false

    解题思路:运用DFS对所有结果进行遍历

    class Solution {
        
        private static int[][] next = new int[][]{{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
        
        public boolean exist(char[][] board, String word) {
            char[] ch = word.toCharArray();
            boolean[][] book = new boolean[board.length][board[0].length];
            for(int i = 0; i < board.length; i++) {
                for(int j = 0; j < board[i].length; j++) {
                    if(check(board, ch, book, i, j, 0)) {
                        return true;
                    }
                }
            }
            
            return false;
        }
        
        private boolean check(char[][] board, char[] ch, boolean[][] book, int i, int j, int cur) {
            if(board[i][j] != ch[cur]) {
                return false;
            } else if(cur >= ch.length - 1) {
                return true;
            }
            book[i][j] = true;
            for(int k = 0; k < 4; k++) {
                int next_i = i + next[k][0], next_j = j + next[k][1];
                if(next_i >= 0 && next_i < board.length && next_j >= 0 && next_j < board[0].length && !book[next_i][next_j]) {
                    book[next_i][next_j] = true;
                    if(check(board, ch, book, next_i, next_j, cur + 1)) {
                        return true;
                    }
                    book[next_i][next_j] = false;
                }
            }
            book[i][j] = false;
            return false;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/katoMegumi/p/14536571.html
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