理解错题了,就是让你在矩阵中循环判断而已,我还想到了康托付展开
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:
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(1-9)或者 '.'
来源:力扣(LeetCode)
链接:https://leetcode.cn/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。
/** * @param {character[][]} board * @return {boolean} */ var isValidSudoku = function(board) { // 理解错题了,就是让你在矩阵中循环判断而已 const rectArr=[] for(let i=0;i<9;i++){ rectArr.push([i,i+1,0,9]) rectArr.push([0,9,i,i+1]) } for(let i=0;i<9;i=i+3){ for(let j=0;j<9;j=j+3){ rectArr.push([i,i+3,j,j+3]) } } console.log(rectArr) let isOk=true; for(let i=0;i<rectArr.length;i++){ if(!isOk){ break; } const arr=[] const rect=rectArr[i]; for(let j=rect[0];j<rect[1];j++){ for(let k=rect[2];k<rect[3];k++){ const t=board[j][k]; if(t!=='.'){ if(arr.indexOf(t)===-1){ arr.push(t) }else{ isOk=false; break } } } } } return isOk };