Leetcode算法Java全解答--36. 有效的数独.md

Posted by lizhao on 07-09,2019

Leetcode算法Java全解答--36. 有效的数独

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题目

判断一个 9x9 的数独是否有效。只需要根据以下规则,验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。 数字 1-9 在每一列只能出现一次。 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字,空白格用 '.' 表示。 说明:

一个有效的数独(部分已被填充)不一定是可解的。 只需要根据以上规则,验证已经填入的数字是否有效即可。 给定数独序列只包含数字 1-9 和字符 '.' 。 给定数独永远是 9x9 形式的。

示例 1:

输入:
[
  ["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"]
]
输出: true
示例 2:

输入:
[
  ["8","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"]
]
输出: false
解释: 除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。
     但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
说明:

一个有效的数独(部分已被填充)不一定是可解的。
只需要根据以上规则,验证已经填入的数字是否有效即可。
给定数独序列只包含数字 1-9 和字符 '.' 。
给定数独永远是 9x9 形式的。

想法

// TODO

结果

// TODO

总结

// TODO

代码

我的答案

	public boolean isValidSudoku(char[][] board) {
		for (int i = 0; i < 9; i++) {
			HashSet<Character> row = new HashSet<>();
			HashSet<Character> column = new HashSet<>();
			HashSet<Character> cube = new HashSet<>();
			for (int j = 0; j < 9; j++) {
				// 检查第i行,在横坐标位置
				if (board[i][j] != '.' && !row.add(board[i][j]))
					return false;
				// 检查第i列,在纵坐标位置
				if (board[j][i] != '.' && !column.add(board[j][i]))
					return false;
				// 行号+偏移量
				int RowIndex = 3 * (i / 3) + j / 3;
				// 列号+偏移量
				int ColIndex = 3 * (i % 3) + j % 3;
				//每个小九宫格,总共9个
				if (board[RowIndex][ColIndex] != '.' 
						&& !cube.add(board[RowIndex][ColIndex]))
					return false;
			}
		}
		return true;
	}

  

大佬们的答案

 // 使用位图法,出现置为1
	public boolean isValidSudoku(char[][] board) {
		for (int i = 0; i < 9; i++) {
			int[] bit_map_row = new int[9];
			int[] bit_map_col = new int[9];
			int[] bit_map_cube = new int[9];
			// 注意值减去1,下标从0开始
			for (int j = 0; j < 9; j++) {
				if (board[i][j] != '.')
					if (bit_map_row[board[i][j] - '1'] == 1)
						return false;
					else
						bit_map_row[board[i][j] - '1'] = 1;
				if (board[j][i] != '.')
					if (bit_map_col[board[j][i] - '1'] == 1)
						return false;
					else
						bit_map_col[board[j][i] - '1'] = 1;
				int RowIndex = 3 * (i / 3) + j / 3;
				int ColIndex = 3 * (i % 3) + j % 3;
				int val = board[RowIndex][ColIndex];
				if (val != '.')
					if (bit_map_cube[val - '1'] == 1)
						return false;
					else
						bit_map_cube[val - '1'] = 1;
			}
		}
		return true;
	}

测试用例


其他

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