# 036 -- 有效的数独
# 题目
题目
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。
- 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
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- 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。
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# 解析 解法说明
提示
🔊 解决这样的问题,肯定是需要遍历数独的二维数组,按照暴力解法,我们需要遍历数组三遍, 每一遍去检查一组数据的正确性,那算法肯定是很慢的,这里我们可以用空间换取时间,利用 hash表记录遍历过的数据,当后续的数据在hash表中发现已经存在了,说明数独是无效的,当 循环结束时,我们就能认为该数独是有效的。
- 我们需要检查三组数据,行数据、列数据、方块数据
- 我们可以为每组数据创建一个数组,其中存储hash表
- 遍历数独数据,每次都去三个数组对应的hash表中查找是否存在该数据
- 存在,则数独无效,不存在,则设置 num:1 的形式
- 循环如果执行完成,则说明数独是有效的。
# 题解源码
/**
* @description
* 创建三组数据,分别存储行数据、列数据、方块容器
* 每个容器的对应数组中存储着hash表
* 每次都将这次遍历的数据存储到对应容器的对应hash中,均标记为1
* {3:1,5:1,...}
* 当在某个容器中发现了对应的key值为1,则说明该数字在该容器中重复
* 则不是数独数组,返回false
* 循环结束后,返回true
* @param {character[][]} board
* @return {boolean}
*/
var isValidSudoku = function (board) {
// 创造行容器、列容器、方块容器
let rows = [];
let cols = [];
let boxes = [];
// i => rows j => cols
for (let i = 0; i < 9; i++) {
// 初始化行map
rows[i] = {};
for (let j = 0; j < 9; j++) {
let key = board[i][j];
// 行数据
if (rows[i][key] && key !== '.') {
return false;
} else {
rows[i][key] = 1;
}
// 列数据
cols[j] = cols[j] || {};
if (cols[j][key] && key !== '.') {
return false;
} else {
cols[j][key] = 1;
}
// 方格数据
// boxes的index,三种写法均为先下取整
let boxIndex = ((i / 3) | 0) * 3 + (j / 3) | 0;
// let boxIndex = ((~~(i / 3)) * 3) + (~~(j / 3));
// let boxIndex = Math.floor(i / 3) * 3 + Math.floor(j / 3);
console.log(boxIndex);
boxes[boxIndex] = boxes[boxIndex] || {};
if (boxes[boxIndex][key] && key !== '.') {
return false
} else {
boxes[boxIndex][key] = 1;
}
}
}
return true;
};
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