Leetcode36. 有效的数独

Leetcode36. 有效的数独

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。
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

题解：
可以使用 box_index = (row / 3) * 3 + columns / 3，其中 / 是整数除法。

scala代码：

/**
    *
    * @param board
    * @return
    */
  def isValidSudoku(board: Array[Array[Char]]): Boolean = {
    val row = new Array[mutable.HashMap[Int, Int]](9)
    val col = new Array[mutable.HashMap[Int, Int]](9)
    val box = new Array[mutable.HashMap[Int, Int]](9)
    for(i<-0 until 9){
        row(i)= new mutable.HashMap[Int,Int]()
        col(i)= new mutable.HashMap[Int,Int]()
        box(i)= new mutable.HashMap[Int,Int]()
      }

    var flag = true
    for (i <- 0 until 9; if (flag)) {
      for (j <- 0 until 9; if (flag)) {
        val char = board(i)(j)
        if (char != '.') {
          val num: Int = char-'0'

          val box_index = (i / 3) * 3 + (j / 3)
          row(i).put(num, row(i).getOrElse(num, 0) + 1)
          col(j).put(num, col(j).getOrElse(num, 0) + 1)
          box(box_index).put(num, box(box_index).getOrElse(num, 0) + 1)
          if (row(i).getOrElse(num, 0) > 1 || col(j).getOrElse(num, 0) > 1 || box(box_index).getOrElse(num, 0) > 1) {
            flag = false
          }
        }
      }
    }
    flag
  }

来源：CSDN

作者：会流泪de鱼

链接：https://blog.csdn.net/sunhaiting666/article/details/104837000