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題目:valid-sudoku(有效的數獨)java
描述:判斷一個 9x9 的數獨是否有效。只須要根據如下規則,驗證已經填入的數字是否有效便可。linux
上圖是一個部分填充的有效的數獨。算法
數獨部分空格內已填入了數字,空白格用 '.'
表示。服務器
示例 1:ide
輸入:
[
["5","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]
]post
輸出: true學習
說明:url
解題過程: spa
利用 set 不重複保存的特性,將數獨元素存入set中,若是不能保存,就判斷爲無效。
精髓是給數字按不一樣維度標記。由於要數字不重複, 那麼能夠給每一個數字在 行、列、九宮格 裏 分別給到不一樣的標識,好比
解法:
1 import java.util.HashSet; 2 import java.util.Set; 3 4 /** 5 * description : https://leetcode.com/problems/valid-sudoku/ 6 * 題目描述 : https://leetcode-cn.com/problems/valid-sudoku/ 7 */ 8 public class ValidSudoku { 9 10 public boolean isValidSudoku(char[][] board) { 11 Set seen = new HashSet(); 12 for (int i = 0; i < 9; ++i) { 13 for (int j = 0; j < 9; ++j) { 14 char number = board[i][j]; 15 if (number != '.') 16 if (!seen.add(number + " in row " + i) || 17 !seen.add(number + " in column " + j) || 18 !seen.add(number + " in block " + i / 3 + "-" + j / 3)) 19 return false; 20 } 21 } 22 return true; 23 } 24 }
題目:
如何在linux服務器 後臺運行一個程序?
參考這篇文章 :Linux命令之nohup 和 重定向