LeetCode中有效的数独题解

LeetCode中有效的数独题解-java

题目

36. 有效的数独

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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。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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。

题解

方法一：

思路

要求行，列，方块中不能有重复的数字，采用Boolean二维数组来保存是否出现此数字，新建三个Boolean二维数组row，col,block,默认值为false，表示全未出现过，然后遍历二维数组board，当此值不等于‘.’时，首先取出此下标的值，并计算对应的下标，然后根据遍历的行号列好计算对应的块号，然后判断是否在row，col,block数组对应下标的值是否为true（为true表示之前已出现过此数字，重复），如果有一个及以上为true，则返回false，如果全为false，则将row，col,block数组对应的下标的值设置为true，代表此数字已出现，下次不能再出现，若遍历结束，仍未发现有重复的数字（满足要求），则返回true

代码实现

class Solution {
    public boolean isValidSudoku(char[][] board) {

        // 记录某行中某位数字是否已出现
        boolean[][] row = new boolean[9][9];
        // 记录某列中某位数字是否已出现
        boolean[][] col = new boolean[9][9];
        // 记录某3*3宫格内某位数字是否已出现
        boolean[][] block = new boolean[9][9];

        // 遍历board二维数组
        for(int i = 0; i < 9; i++){
            for(int j = 0; j < 9; j++){
                // 当board[i][j] 不等于‘.’ 时，计算此时值对应的下标与此时值所在的块号坐标
                if(board[i][j] != '.'){
                    // 纵坐标长度为9，分别对应1-9个数字，其下标对应为数字字符-1字符
                    int num = board[i][j] -'1';
                    // 计算对应块号的下标
                    int blockIndex = i /3 * 3 + j / 3;
                    // 判断是否满足行，列与3*3宫格中是否出现，
                    if(row[i][num] || col[j][num] || block[blockIndex][num]){
                        return false;
                    }else{
                        // 若未出现，不返回false，并将行，列与此块中对应的下标值设为true
                        row[i][num] = true;
                        col[j][num] = true;
                        block[blockIndex][num] = true;
                    }
                }
            }
        }
        return true;
    }
}

复杂度分析

• 时间复杂度：O(1)，因为board为9*9数组，n为常数，所以时间复杂度为O(1)
• 空间复杂度：O(1)，使用了额外固定大小的Boolean二维数组，空间复杂度为O(1)

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方法二：

思路

代码实现

复杂度分析

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