一、題目

判斷一個(gè) 9x9 的數(shù)獨(dú)是否有效。只需要根據(jù)以下規(guī)則，驗(yàn)證已經(jīng)填入的數(shù)字是否有效即可。

1. 數(shù)字 1-9 在每一行只能出現(xiàn)一次。
2. 數(shù)字 1-9 在每一列只能出現(xiàn)一次。
3. 數(shù)字 1-9 在每一個(gè)以粗實(shí)線(xiàn)分隔的 3x3 宮內(nèi)只能出現(xiàn)一次。

image

<small style="box-sizing: border-box; font-size: 10.399999618530273px;">上圖是一個(gè)部分填充的有效的數(shù)獨(dú)。</small>

數(shù)獨(dú)部分空格內(nèi)已填入了數(shù)字，空白格用 '.' 表示。

示例 1:

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"]
]
輸出: true

示例 2:

輸入:
[
  ["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"]
]
輸出: false
解釋: 除了第一行的第一個(gè)數(shù)字從 5 改為 8 以外，空格內(nèi)其他數(shù)字均與 示例1 相同。
     但由于位于左上角的 3x3 宮內(nèi)有兩個(gè) 8 存在, 因此這個(gè)數(shù)獨(dú)是無(wú)效的。

說(shuō)明:

• 一個(gè)有效的數(shù)獨(dú)（部分已被填充）不一定是可解的。
• 只需要根據(jù)以上規(guī)則，驗(yàn)證已經(jīng)填入的數(shù)字是否有效即可。
• 給定數(shù)獨(dú)序列只包含數(shù)字 1-9 和字符 '.' 。
• 給定數(shù)獨(dú)永遠(yuǎn)是 9x9 形式的。

二、解題

直接遍歷二維數(shù)組，在第二個(gè)for循環(huán)里判斷每個(gè)元素，在其對(duì)應(yīng)的行、列已經(jīng)子數(shù)獨(dú)里驗(yàn)證是否有重復(fù)數(shù)字即可。
先創(chuàng)建三個(gè)字典rowDict:[Int:[Character:Int]]（存儲(chǔ)所有行的數(shù)字），columnDict: [Int:[Character:Int]]（存儲(chǔ)所有列的數(shù)字），boxDict: [Int:[Character:Int]]（存儲(chǔ)所有3x3子數(shù)獨(dú)中的數(shù)字）。
這里有個(gè)難點(diǎn)，就是子數(shù)獨(dú)的位置計(jì)算let bIndex = i / 3 * 3 + j / 3。
詳細(xì)請(qǐng)看代碼：
時(shí)間復(fù)雜度：O(m*n)。(假設(shè)board[m][n])

三、代碼實(shí)現(xiàn)

    class Solution {
        func isValidSudoku(_ board: [[Character]]) -> Bool {
            var rowDict: [Int:[Character:Int]] = [:]
            var columnDict: [Int:[Character:Int]] = [:]
            var boxDict: [Int:[Character:Int]] = [:]
            for i in (0..<9) {
                for j in (0..<9) {
                    let c = board[i][j]
                    if c == "." {
                        continue
                    }
                    if rowDict[i] == nil {
                        rowDict[i] = [:]
                    }
                    if rowDict[i]![c] != nil {
                        return false
                    }else{
                        rowDict[i]![c] = 1
                    }
                    
                    if columnDict[j] == nil {
                        columnDict[j] = [:]
                    }
                    if columnDict[j]![c] != nil {
                        return false
                    }else{
                        columnDict[j]![c] = 1
                    }

                    let bIndex = i / 3 * 3 + j / 3
                    if boxDict[bIndex] == nil {
                        boxDict[bIndex] = [:]
                    }
                    if boxDict[bIndex]![c] != nil {
                        return false
                    }else {
                        boxDict[bIndex]![c] = 1
                    }
                }
            }
            return true
        }
    }

Demo地址：github