前端算法题解leetcode36-有效的数独示例
目录
- 题目
- 解题思路-分别处理
- 代码实现
- 解题思路-一次扫描判断所有
- 代码实现
题目
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 '.'
解题思路-分别处理
分别处理每一行、每一列以及每一个九宫格,哪一部分验证失败,都返回 false
,如果都验证通过,返回 true
。
代码实现
function getOrigin(){ return { '1':0, '2':0, '3':0, '4':0, '5':0, '6':0, '7':0, '8':0, '9':0, } } function checkArr(arr){ const counts = getOrigin() for(let i = 0;i<9;i++){ if(counts[arr[i]]){ return false }else{ counts[arr[i]]++ } } return true } var isValidSudoku = function(board) { // 处理每一行 for(let i = 0;i<9;i++){ if(!checkArr(board[i])){ return false } } // 处理每一列 for(let i = 0;i<9;i++){ const arr = [] for(let j = 0;j<9;j++){ if(board[j][i] === '.'){ continue } arr.push(board[j][i]) } if(!checkArr(arr)){ return false } } // 处理 9 个九宫格 for(let i = 0;i<3;i++){ for(let j = 0;j<3;j++){ const arr = [] for(let k = j*3;k<j*3+3;k++){ for(let h = 3*i;h<3*i+3;h++){ if(board[k][h] === '.'){ continue } arr.push(board[k][h]) } } if(!checkArr(arr)){ return false } } } return true }
解题思路-一次扫描判断所有
首先创建 lines
记录每一行出现的数字的次数,columns
记录每一列出现的数字的次数,scratchableLatexs
记录每一个九空格出现的数字的次数。
然后双层循环可以遍历输入数组中的每一个元素,根据当前 i
,j
值判断属于哪一行,哪一列以及哪一个九宫格,分别判断即可。
代码实现
function getOrigin(){ return { '1':0, '2':0, '3':0, '4':0, '5':0, '6':0, '7':0, '8':0, '9':0, } } var isValidSudoku = function(board) { const lines = [] const columns = [] const scratchableLatexs = [] for(let i = 0;i<9;i++){ lines[i] = getOrigin() columns[i] = getOrigin() scratchableLatexs[i] = getOrigin() } for(let i = 0;i<9;i++){ for(let j = 0;j<9;j++){ const item = board[i][j] if(item === '.'){ continue } if(lines[i][item]){ return false }else{ lines[i][item]++ } if(columns[j][item]){ return false }else{ columns[j][item]++ } if(i<3){ if(j<3){ if(scratchableLatexs[0][item]){ return false }else{ scratchableLatexs[0][item]++ } }else if(j<6){ if(scratchableLatexs[1][item]){ return false }else{ scratchableLatexs[1][item]++ } }else{ if(scratchableLatexs[2][item]){ return false }else{ scratchableLatexs[2][item]++ } } }else if(i<6){ if(j<3){ if(scratchableLatexs[3][item]){ return false }else{ scratchableLatexs[3][item]++ } }else if(j<6){ if(scratchableLatexs[4][item]){ return false }else{ scratchableLatexs[4][item]++ } }else{ if(scratchableLatexs[5][item]){ return false }else{ scratchableLatexs[5][item]++ } } }else{ if(j<3){ if(scratchableLatexs[6][item]){ return false }else{ scratchableLatexs[6][item]++ } }else if(j<6){ if(scratchableLatexs[7][item]){ return false }else{ scratchableLatexs[7][item]++ } }else{ if(scratchableLatexs[8][item]){ return false }else{ scratchableLatexs[8][item]++ } } } } } return true }
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