ARTS 打卡 第 30 周

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
提示：
board 和 word 中只包含大写和小写英文字母。1 <= board.length <= 2001 <= board[i].length <= 2001 <= word.length <= 10^3

class Solution {    /**     * 对当前节点四个方向的模拟 是一个非常有用的技巧     */    private static int[][] directions = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };
    public boolean exist(char[][] board, String word) {        if (board == null || board.length == 0 || board[0].length == 0 || word == null) {            return false;        }        int rowLen = board.length;        int colLen = board[0].length;        boolean[][] visited = new boolean[rowLen][colLen];        for (int row = 0; row < rowLen; row++) {            for (int col = 0; col < colLen; col++) {                if (dnf(board, visited, row, col, word, 0)) {                    return true;                }            }        }        return false;    }
    /**     * board[row][col]为起点     *     * @param board     * @param visited     * @param row     * @param col     * @param word     * @param indexOfWord     * @return     */    private boolean dnf(char[][] board, boolean[][] visited, int row, int col, String word,            int indexOfWord) {        if (board[row][col] != word.charAt(indexOfWord)) {            return false;        }        if (indexOfWord == word.length() - 1) {            return true;        }        visited[row][col] = true;
        boolean result = false;        int rowLen = board.length;        int colLen = board[0].length;        for (int[] dir : directions) {            int nextRow = row + dir[0];            int nextCol = col + dir[1];            if (nextRow >= 0 && nextRow < rowLen && nextCol >= 0 && nextCol < colLen) {                if (!visited[nextRow][nextCol]) {                    boolean flag = dnf(board, visited, nextRow, nextCol, word, indexOfWord + 1);                    if (flag) {                        result = true;                        break;                    }                }            }        }        visited[row][col] = false;        return result;    }
}

Note that “ps -aux” is distinct from “ps aux”. The POSIX and UNIX standards require that “ps -aux” print all processes owned by a user named “x”, as well as printing all processes that would be selected by the -a option. If the user named “x” does not exist, this ps may interpret the command as “ps aux” instead and print a warning.This behavior is intended to aid in transitioning old scripts and habits. It is fragile, subject to change, and thus should not be relied upon.大致意思是：请注意"ps -aux"不同于"ps aux"。POSIX和UNIX的标准要求"ps -aux"打印用户名为"x"的用户的所有进程，以及打印所有将由-a选项选择的过程。如果用户名为"x"不存在，ps的将会解释为"ps aux"，而且会打印一个警告。这种行为是为了帮助转换旧脚本和习惯。它是脆弱的，即将更改，因此不应依赖。

ps #默认情况    PID TTY          TIME CMD1216874 pts/0    00:00:01 zsh1280028 pts/0    00:00:00 ps

#显示用户进程 标准风格:  ps -e  ps -ef  ps -eF  ps -ely
 #显示用户进程  BSD 风格:  ps ax  ps axu