zoukankan      html  css  js  c++  java
  • LeetCode

    题目:

    Given a 2D board and a word, find if the word exists in the grid.

    The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

    For example,
    Given board =

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

    word = "ABCCED", -> returns true,
    word = "SEE", -> returns true,
    word = "ABCB", -> returns false.

    思路:

    递归;及时return true,减少计算量

    package recursion;
    
    public class WordSearch {
    
        public boolean exist(char[][] board, String word) {
            int m;
            int n;
            int wordLen;
            if (board == null || word == null || 
                    (m = board.length) == 0 || (n = board[0].length) == 0 || 
                    (wordLen = word.length()) == 0 || wordLen > m * n) 
                return false;
            boolean[][] visited = new boolean[m][n];
    
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (dfs(board, i, j, visited, word, 0, m, n, wordLen))
                        return true;
                }
            }
            return false;
        }
     
        private boolean dfs(char[][] board, int x, int y, boolean[][] visited, String word, int pos, int m, int n, int wordLen) {
            if (visited[x][y] || board[x][y] != word.charAt(pos)) return false;
            if (pos == wordLen - 1) return true;
            visited[x][y] = true;
    
            if (x - 1 >= 0 && dfs(board, x - 1, y, visited, word, pos + 1, m, n, wordLen)) return true;
            
            if (x + 1 < m && dfs(board, x + 1, y, visited, word, pos + 1, m, n, wordLen)) return true;
            
            if (y - 1 >= 0 && dfs(board, x, y - 1, visited, word, pos + 1, m, n, wordLen)) return true;
            
            if (y + 1 < n && dfs(board, x, y + 1, visited, word, pos + 1, m, n, wordLen)) return true;
            
            visited[x][y] = false;
            
            return false;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            char[][] board = 
                            {
                                {'A','B','C','E'},
                                {'S','F','C','S'},
                                {'A','D','E','E'}
                            };
            WordSearch w = new WordSearch();
            System.out.println(w.exist(board, "ABCCED"));
            System.out.println(w.exist(board, "SEE"));
            System.out.println(w.exist(board, "ABCB"));
        }
    
    }
  • 相关阅读:
    多线程:多线程设计模式(一):总体介绍
    javascript:12种JavaScript MVC框架之比较
    mysql 查询死锁语句
    charles 抓包工具破解方法
    java 自定义log类
    git统计日期之间的代码改动行数
    mac/linux自带定时任务执行crontab的使用
    python MD5步骤
    python 操作excel读写
    python logger日志工具类
  • 原文地址:https://www.cnblogs.com/null00/p/5094747.html
Copyright © 2011-2022 走看看