LeetCode题目36：有效的数独

原题描述

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

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。

原题链接：https://leetcode-cn.com/problems/valid-sudoku

思路解析

数独确实有一道难题，但不是这一道，此题反而属于一点就透的类型。

回到问题，要判断是否有重复的数组出现，也就是判断重复，根据之前的经验，选择hash table绝对错不了，对于规模固定为9*9的数独来说，这点存储空间的浪费不算什么。

先考虑行。要判断某一行是否有重复的数字，我们只需要遍历这一行，统计每个数字出现的次数即可。一共9行，那么就需要9个这样的hash table；如果开辟一个长度为9的hash table的数组，那么行号和数组index就可以一一对应起来了。

列也是如此，也需要一个长度为9的hash table数组。

3*3子数独也需要长度为9的hash table。那么给定一个二维坐标(x,y)，如何判断它属于第几个子数独？假设我们如下编号，那么(x, y)和子数独index的关系是：

index = (x / 3) * 3 + y / 3

我们可以一边扫描数独，一边将统计信息填入这三类hash table中，然后再检查是否有某个数字出现的次数多于1即可。最多扫描一遍，就可以判断出结果。

复杂度分析

时间复杂度：O(1)
空间复杂度：O(1)

C++参考代码

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        std::vector<std::unordered_map<char, int>> rows(9);
        std::vector<std::unordered_map<char, int>> cols(9);
        std::vector<std::unordered_map<char, int>> box(9);

        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if (board[i][j] != '.') {
                    int num = (int)board[i][j];
                    int bid = (i / 3) * 3 + j / 3;
                    rows[i][num] = rows[i].count(num) != 0 ? rows[i][num] + 1 : 1;
                    cols[j][num] = cols[j].count(num) != 0 ? cols[j][num] + 1 : 1;
                    box[bid][num] = box[bid].count(num) != 0 ? box[bid][num] + 1 : 1;
                    
                    if (rows[i][num] > 1 || cols[j][num] > 1 || box[bid][num] > 1) {
                        return false;
                    }
                }
            }
        }

        return true;
    }
};