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有效的数独
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。
方法
思路：
- 构造一个包含 每一行 每一列 以及 3x3 宫 的数组，构造数组时将空格即'.'去掉。
- 判断数组中是否都满足去重后的长度等于原始数组的长度。「判断是否有重复值」
代码：
/**
* @param {character[][]} board
* @return {boolean}
*/
var isValidSudoku = function(board) {
let res = []
const gridRes = {
'00': [],
'01': [],
'02': [],
'10': [],
'11': [],
'12': [],
'20': [],
'21': [],
'22': []
}
for(let i=0;i<board.length;i++) {
res.push(board[i].filter((item)=>item!=='.'))
const rowRes = []
for(let j=0;j<board[i].length;j++) {
if(board[j][i] !== '.'){
rowRes.push(board[j][i])
}
if(board[i][j] !== '.'{
gridRes[`${Math.floor(i/3)}${Math.floor(j/3)}`].push(board[i][j])
}
}
res.push(rowRes)
}
res = res.concat(Object.values(gridRes))
return res.every((item=>{
return Array.from(new Set(item)).length === item.length
}))
};
结果：
- 执行结果： 通过
- 执行用时：76 ms, 在所有 JavaScript 提交中击败了95.20%的用户
- 内存消耗：45.5 MB, 在所有 JavaScript 提交中击败了13.49%的用户
- 通过测试用例：507 / 507
矩阵置零
题目
给定一个 m x n 的矩阵，如果一个元素为 0 ，则将其所在行和列的所有元素都设为 0 。请使用 原地 算法。
示例 1：
输入：matrix = [[1,1,1],[1,0,1],[1,1,1]]
输出：[[1,0,1],[0,0,0],[1,0,1]]
示例 2：
输入：matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
输出：[[0,0,0,0],[0,4,5,0],[0,3,1,0]]
解题方法
思路：
- 遍历数组，将元素为
0的所在行列标记下来，分别为row[i]以及col[j] - 第二次遍历数组，将
row[i]或col[j]为true的matrix[i][j]置为0
代码：
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var setZeroes = function(matrix) {
const row = {}
const col = {}
for(let i=0;i<matrix.length;i++) {
for(let j=0;j<matrix[i].length;j++) {
if(matrix[i][j] === 0) {
row[i] = true
col[j] = true
}
}
}
for(let i=0;i<matrix.length;i++) {
for(let j=0;j<matrix[i].length;j++) {
if(row[i] || col[j]) {
matrix[i][j] = 0
}
}
}
};
结果：
- 执行结果： 通过
- 执行用时：76 ms, 在所有 JavaScript 提交中击败了94.43%的用户
- 内存消耗：44.1 MB, 在所有 JavaScript 提交中击败了9.37%的用户
- 通过测试用例：164 / 164
