一、题目
请你判断一个 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。(请参考示例图)
数独部分空格内已填入了数字,空白格用 ‘.’ 表示。
注意:
- 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。
示例 1:
输入:board =
[["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:board =
[["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。
提示:
- board.length == 9
- board[i].length == 9
- board[i][j] 是一位数字或者 ‘.’
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/valid-sudoku
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二、分析及代码
1. 模拟
(1)思路
设计布尔变量,记录各行、列、3x3 格子内各个数字的出现情况。遍历数组中的元素,根据数字在三种限定范围内的重复情况,判断数独有效性。
(2)代码
class Solution {
public boolean isValidSudoku(char[][] board) {
//初始化
boolean[][] rows = new boolean[9][9];//rows[i][j]表示第i行数字j是否已出现过
boolean[][] cols = new boolean[9][9];//cols[i][j]表示第i列数字j是否已出现过
boolean[][][] boxes = new boolean[3][3][9];//boxes[i][j][k]表示对于3x3的格子第i行第j个中数字k是否已出现过
for (int i = 0; i < 9; i++)
for (int j = 0; j < 9; j++) {
rows[i][j] = false;
cols[i][j] = false;
}
for (int i = 0 ; i < 3; i++)
for (int j = 0; j < 3; j++)
for (int k = 0; k < 9; k++)
boxes[i][j][k] = false;
//遍历验证
for (int i = 0; i < 9; i++) {
//遍历行
for (int j = 0; j < 9; j++) {
//遍历列
if (board[i][j] != '.') {
//判断数字是否已填入
int num = board[i][j] - '1';//提取数字-1的值
if (rows[i][num] == true || cols[j][num] == true || boxes[i / 3][j / 3][num] == true)//数字重复,数独无效
return false;
rows[i][num] = true;//更新数字出现标记
cols[j][num] = true;//更新数字出现标记
boxes[i / 3][j / 3][num] = true;//更新数字出现标记
}
}
}
return true;//无重复数字,数独有效
}
}
(3)结果
执行用时 :2 ms,在所有 Java 提交中击败了 70.66% 的用户;
内存消耗 :38.5 MB,在所有 Java 提交中击败了 42.87% 的用户。
三、其他
暂无。