有效的数独
请你判断一个 9 x 9 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）
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。 空白格用 '.' 表示。
示例 1：
codeType
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：
codeType
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（1-9）或者 '.'
解法：
ts
function isValidSudoku(board: string[][]): boolean {
// 暴力法就是对每个数组的行和列进行检测
// 对于这种存在大量重复检测的并且是O(n)的题，可以用hashmap存储检测过的内容
// 从而使查找时间变成O(1)
// 每行、每列、每九宫格各一个hashmap
const n = board[0].length
const rowMaps = new Array()
const colMaps = new Array()
const neibMaps = new Array()
for(let i=0; i<n; i++){
const rowMap = new Map()
const colMap = new Map()
const neibMap = new Map()
rowMaps.push(rowMap)
colMaps.push(colMap)
neibMaps.push(neibMap)
}
// 按行遍历
for(let i=0; i<n; i++){
for(let j=0; j<n; j++){
const curNum = board[i][j]
// 跳过'.'
if(curNum === '.') continue
// 检查行
if(rowMaps[i].has(curNum)){
return false
}else{
rowMaps[i].set(curNum, true)
}
// 检查列
if(colMaps[j].has(curNum)){
return false
}else{
colMaps[j].set(curNum, true)
}
// 检查九宫格(0~8)
const neibID = Math.floor(i / 3) * 3 + Math.floor(j / 3)
if(neibMaps[neibID].has(curNum)){
return false
}else{
neibMaps[neibID].set(curNum, true)
}
}
}
return true
};