LeetCode_Medium_36. Valid Sudoku

2019.2.13

题目描述：

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

Example 1:

Input:
[
  ["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"]
]
Output: true

Example 2:

Input:
[
  ["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"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:

A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.
The given board contain only digits 1-9 and the character '.'.
The given board size is always 9x9.

验证一个9x9的序列是不是数独的类型，首先属于数独的要求是每行，每列，每个3x3的小方阵内的数字都是唯一的。其实这个题目的逻辑算法很简单，就是行、列包括小方阵都要进行判断即可。

解法一：

我们需要三个标志矩阵，分别记录各行，各列，各小方阵是否出现某个数字，其中行和列标志下标很好对应，就是小方阵的下标需要稍稍转换一下

C++代码：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int row[10][10]={0};
        int col[10][10]={0};
        int cell[10][10]={0};
        for(int i=0;i<9;++i){
            for(int j=0;j<9;++j){
                char& ch=board[i][j];
                if(ch=='.') continue;
                int r1=++row[i][ch-'0'];
                int r2=++col[j][ch-'0'];
                int r3=++cell[i/3*3+j/3][ch-'0'];
                if(r1>1 || r2>1 || r3>1) return false;
            }
        }
        return true;
    }
};

解法二：官方解法

其实思想都一样，可能解释的更好，我就贴出来了

Java代码：

class Solution {
  public boolean isValidSudoku(char[][] board) {
    // init data
    HashMap<Integer, Integer> [] rows = new HashMap[9];
    HashMap<Integer, Integer> [] columns = new HashMap[9];
    HashMap<Integer, Integer> [] boxes = new HashMap[9];
    for (int i = 0; i < 9; i++) {
      rows[i] = new HashMap<Integer, Integer>();
      columns[i] = new HashMap<Integer, Integer>();
      boxes[i] = new HashMap<Integer, Integer>();
    }

    // validate a board
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) {
        char num = board[i][j];
        if (num != '.') {
          int n = (int)num;
          int box_index = (i / 3 ) * 3 + j / 3;

          // keep the current cell value
          rows[i].put(n, rows[i].getOrDefault(n, 0) + 1);
          columns[j].put(n, columns[j].getOrDefault(n, 0) + 1);
          boxes[box_index].put(n, boxes[box_index].getOrDefault(n, 0) + 1);

          // check if this value has been already seen before
          if (rows[i].get(n) > 1 || columns[j].get(n) > 1 || boxes[box_index].get(n) > 1)
            return false;
        }
      }
    }

    return true;
  }
}

LeetCode_Medium_36. Valid Sudoku

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