前端算法题解leetcode36-有效的数独示例
作者：前端_奔跑的蜗牛
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
}
至此我们就完成了 leetcode-36-有效的数独，更多关于前端算法题解有效的数独的资料请关注脚本之家其它相关文章！