Leetcode数据结构题目集第五天，有效的数独和矩阵置零.
其中矩阵置零比较有趣放在前面，有效的数独只是代码量比较长，逻辑部分并不难。

矩阵置零

给定一个 m x n 的矩阵，如果一个元素为 0 ，则将其所在行和列的所有元素都设为 0 。请使用原地算法。

进阶：

  • 一个直观的解决方案是使用  O(mn) 的额外空间，但这并不是一个好的解决方案。
  • 一个简单的改进方案是使用 O(m + n) 的额外空间，但这仍然不是最好的解决方案。
    你能想出一个仅使用常量空间的解决方案吗？

 

示例 1：

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输入：matrix = [[1,1,1],[1,0,1],[1,1,1]]
输出：[[1,0,1],[0,0,0],[1,0,1]]

示例 2：

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输入：matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
输出：[[0,0,0,0],[0,4,5,0],[0,3,1,0]]

提示：

  • m == matrix.length
  • n == matrix[0].length
  • 1 <= m, n <= 200
  • -231 <= matrix[i][j] <= 231 - 1

解题思路&题解

通过观察发现置零是将0的行号和列号记录下来，然后将记录下来的行列置零即可
具体思路如下：

  • 当同一行有一个0时，则该行后面遇到0值时也无需关系这一行的变化->因为一行里面有一个0，整行都是0（列同理）
  • 在同一行有两个0时需要注意，这意味着有两列也需要置零。

代码如下：

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func setZeroes(matrix [][]int) {
	row, col := []int{}, []int{}
	for r := 0; r < len(matrix); r++ {
		for c := 0; c < len(matrix[r]); c++ {
			if matrix[r][c] == 0 {
				// 记录
				row = append(row, r)
				col = append(col, c)
			}
		}
	}
    
	for _, item := range row {
		for index := range matrix[item] {
			matrix[item][index] = 0
		}
	}

	for _, item := range col {
		for _, value := range matrix {
			value[item] = 0
		}
	}
}

有效的数独

请你判断一个 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。（请参考示例图）
数独部分空格内已填入了数字，空白格用 ‘.’ 表示。

注意：

  • 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。

示例 1：

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输入：board = 
[["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：

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输入：board = 
[["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。

提示：

  • board.length == 9
  • board[i].length == 9
  • board[i][j] 是一位数字或者 ‘.’

解题思路&题解

题目没什么难度，唯一需要注意的是：判断子矩阵的时候，是根据当前坐标/矩阵大小来决定的。
代码如下：

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func isValidSudoku(board [][]byte) bool {
	rowBuf := make([][9]bool, 9)
	colBuf := make([][9]bool, 9)
	boxBuf := make([][9]bool, 9)
	
	for row:=0; row<9; row++ {
		for col:=0; col<9; col++ {
			if board[row][col] == '.' {
				continue
			}
			
			val := board[row][col] - '1'
			if rowBuf[row][val] || colBuf[col][val] || boxBuf[row/3*3+col/3][val] {
				return false
			}
			
			rowBuf[row][val] = true
			colBuf[col][val] = true
			boxBuf[row/3*3+col/3][val] = true
		}
	}
	return true
}

func validDigit(digit byte) bool {
	return digit >= '1' && digit <= '9'
}

func inRow(rows []byte, col int, digit byte) bool {
	for c := 0; c < 9; c++ {
		if rows[c] == digit && c != col {
			return true
		}
	}
	return false
}

func inColumn(g [][]byte, column, row int, digit byte) bool {
	for r := 0; r < 9; r++ {
		if g[r][column] == digit && r != row {
			return true
		}
	}
	return false
}

func inRegion(g [][]byte, row, column int, digit byte) bool {
	startRow, startColumn := row/3*3, column/3*3
	for r := startRow; r < startRow+3; r++ {
		for c := startColumn; c < startColumn+3; c++ {
			if g[r][c] == digit && r != row && c != column {
				return true
			}
		}
	}
	return false
}