标签：空格 res 商业 i++ 字符 链接 个数 open sudoku
36. 有效的数独
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。
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。来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-sudoku
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
解答：
个人思路：
1、检测每行是否有重复数字；
2、检测每列是否有重复数字；
3、检测每个3*3内是否有重复数字。
96ms
/** * @param {character[][]} board * @return {boolean} */ function isRepeat(arr) { let hash = {} for (let i = 0; i < arr.length; i++) { if (!isNaN(arr[i])) { if (hash[arr[i]]) { return false } else { hash[arr[i]] = 1 } } } return true } var isValidSudoku = function (board) { let result let colArr = [] let blockArr = [] //每一行是否重复检测 for (let i = 0; i < board.length; i++) { result = isRepeat(board[i]) if (!result) { console.log(‘行重复‘) return false } } //每一列是否重复检测 for (let i = 0; i < board.length; i++) { for (let j = 0; j < board[i].length; j++) { if (!colArr[j]) { colArr[j] = [] } colArr[j].push(board[i][j]) } } for (let i = 0; i < colArr.length; i++) { result = isRepeat(colArr[i]) if (!result) { console.log(‘列重复‘) return false } } //3*3是否重复 let block = 0 for (let i = 0; i < board.length; i++) { let reset = block for (let j = 0; j < board[i].length; j++) { if (!blockArr[block]) { blockArr[block] = [] } blockArr[block].push(board[i][j]) if ((j + 1) % 3 === 0) { block++ } if (j === 8) { block = reset } } if ((i + 1) % 3 === 0) { block += 3 } } for (let i = 0; i < blockArr.length; i++) { result = isRepeat(blockArr[i]) if (!result) { console.log(‘3*3重复‘) return false } } //以上三种检测都通过 return true };
【Leetcode】【中等】【36. 有效的数独】【JavaScript】
标签：空格 res 商业 i++ 字符 链接 个数 open sudoku
原文地址：https://www.cnblogs.com/2463901520-sunda/p/11620245.html
