36. 有效的数独,37. 解数独,38. 外观数列,每题做详细思路梳理,配套Python&Java双语代码, 2024.03.16 可通过leetcode所有测试用例。
目录
36. 有效的数独
解题思路
完整代码
Java
Python
37. 解数独
解题思路
完整代码
Java
Python
38. 外观数列
解题思路
完整代码
Java
Python
36. 有效的数独
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
- 遍历数独的每一个格子。
- 对于每个非空格子,检查它所在的行、列以及宫内是否有重复的数字。
- 为了方便检查,可以使用三个二维数组分别记录每行、每列和每个宫格中数字的出现情况。例如,
rows[i][num]
表示数字num+1
在第i
行是否已出现,cols[j][num]
表示数字num+1
在第j
列是否已出现,boxes[boxIndex][num]
表示数字num+1
在第boxIndex
个宫内是否已出现。boxIndex
可以通过(i / 3) * 3 + j / 3
计算得到,其中i
和j
分别是格子的行列索引。
完整代码
Java
public class Solution {public boolean isValidSudoku(char[][] board) {boolean[][] rows = new boolean[9][9];boolean[][] cols = new boolean[9][9];boolean[][] boxes = new boolean[9][9];for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {if (board[i][j] != '.') {int num = board[i][j] - '1'; // 将字符转换为数字,并减1以适应数组索引int boxIndex = (i / 3) * 3 + j / 3;if (rows[i][num] || cols[j][num] || boxes[boxIndex][num]) {return false; // 如果在对应的行、列或宫内发现重复,则返回 false}rows[i][num] = true;cols[j][num] = true;boxes[boxIndex][num] = true;}}}return true; // 没有发现重复,返回 true}
}
Python
class Solution:def isValidSudoku(self, board: List[List[str]]) -> bool:rows = [[False] * 9 for _ in range(9)]cols = [[False] * 9 for _ in range(9)]boxes = [[False] * 9 for _ in range(9)]for i in range(9):for j in range(9):if board[i][j] != '.':num = int(board[i][j]) - 1 # 将字符转换为数字,并减1以适应数组索引boxIndex = (i // 3) * 3 + j // 3if rows[i][num] or cols[j][num] or boxes[boxIndex][num]:return False # 如果在对应的行、列或宫内发现重复,则返回 Falserows[i][num] = Truecols[j][num] = Trueboxes[boxIndex][num] = Truereturn True # 没有发现重复,返回 True
37. 解数独
编写一个程序,通过填充空格来解决数独问题。
数独的解法需 遵循如下规则:
- 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)数独部分空格内已填入了数字,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:[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]] 解释:输入的数独如上图所示,唯一有效的解决方案如下所示:提示:
board.length == 9
board[i].length == 9
board[i][j]
是一位数字或者'.'
- 题目数据 保证 输入数独仅有一个解
解题思路
解决数独问题通常采用回溯法,这是一种深度优先搜索的算法,适用于解决决策问题。基本思路是尝试填充数独中的每个空格,每次填充时遵循数独的规则,如果在某个位置无法填入有效数字,则回溯到上一个空格,尝试另一个数字。重复这一过程,直到数独被成功填充。
算法步骤如下:
- 从数独的第一个格子开始,按照某种顺序(通常是从左到右,从上到下)遍历每个格子。
- 对于每个空格,尝试填入数字 1 到 9,每尝试一个数字前,检查行、列和所在的 3x3 宫格内是否已经存在该数字。
- 如果一个数字能够被安全地填入(即不违反数独规则),则填入并递归地尝试填写下一个空格。
- 如果当前空格无论填入何种数字都无法使数独有效,或者所有空格都已成功填写,算法结束。在前一种情况下,需要撤销最后的填写并回溯到上一个空格;在后一种情况下,找到了数独的解。
完整代码
Java
public class Solution {public void solveSudoku(char[][] board) {solve(board);}private boolean solve(char[][] board) {for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {if (board[i][j] == '.') {for (char c = '1'; c <= '9'; c++) {if (isValid(board, i, j, c)) {board[i][j] = c;if (solve(board)) {return true;} else {board[i][j] = '.'; // 回溯}}}return false; // 无解,回溯}}}return true; // 找到解}private boolean isValid(char[][] board, int row, int col, char c) {for (int i = 0; i < 9; i++) {if (board[i][col] == c) return false; // 检查列if (board[row][i] == c) return false; // 检查行if (board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] == c) return false; // 检查 3x3 宫}return true;}
}
Python
class Solution:def solveSudoku(self, board: List[List[str]]) -> None:def solve():for i in range(9):for j in range(9):if board[i][j] == '.':for c in '123456789':if isValid(i, j, c):board[i][j] = cif solve():return Trueelse:board[i][j] = '.' # 回溯return False # 无解,回溯return True # 找到解def isValid(i, j, c):for k in range(9):if board[i][k] == c or board[k][j] == c:return False # 检查行和列if board[3 * (i // 3) + k // 3][3 * (j // 3) + k % 3] == c:return False # 检查 3x3 宫return Truesolve()
38. 外观数列
给定一个正整数
n
,输出外观数列的第n
项。「外观数列」是一个整数序列,从数字 1 开始,序列中的每一项都是对前一项的描述。
你可以将其视作是由递归公式定义的数字字符串序列:
countAndSay(1) = "1"
countAndSay(n)
是对countAndSay(n-1)
的描述,然后转换成另一个数字字符串。前五项如下:
1. 1 2. 11 3. 21 4. 1211 5. 111221 第一项是数字 1 描述前一项,这个数是1
即 “ 一 个 1 ”,记作"11"
描述前一项,这个数是11
即 “ 二 个 1 ” ,记作"21"
描述前一项,这个数是21
即 “ 一 个 2 + 一 个 1 ” ,记作 "1211"
描述前一项,这个数是1211
即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ,记作 "111221"
要 描述 一个数字字符串,首先要将字符串分割为 最小 数量的组,每个组都由连续的最多 相同字符 组成。然后对于每个组,先描述字符的数量,然后描述字符,形成一个描述组。要将描述转换为数字字符串,先将每组中的字符数量用数字替换,再将所有描述组连接起来。
例如,数字字符串
"3322251"
的描述如下图:示例 1:
输入:n = 1 输出:"1" 解释:这是一个基本样例。示例 2:
输入:n = 4 输出:"1211" 解释: countAndSay(1) = "1" countAndSay(2) = 读 "1" = 一 个 1 = "11" countAndSay(3) = 读 "11" = 二 个 1 = "21" countAndSay(4) = 读 "21" = 一 个 2 + 一 个 1 = "12" + "11" = "1211"
解题思路
要生成外观数列的第 n
项,我们可以从第 1
项开始,逐步构建直到第 n
项。对于每一项,我们都按照以下步骤来构建:
- 查看前一项的数字字符串。
- 从左到右扫描字符串,对连续的相同数字进行分组。
- 对每个分组,先记录数字的数量(即分组的长度),然后记录数字本身。
- 将上述所有分组的描述连在一起,形成新的数字字符串。
- 重复上述步骤直到达到第
n
项。
完整代码
Java
public class Solution {public String countAndSay(int n) {String s = "1";for (int i = 1; i < n; i++) {StringBuilder sb = new StringBuilder();char c = s.charAt(0);int count = 1;for (int j = 1; j < s.length(); j++) {if (s.charAt(j) == c) {count++; // 相同字符,计数增加} else {sb.append(count).append(c); // 不同字符,记录前一个字符的计数和字符c = s.charAt(j);count = 1; // 重置计数}}sb.append(count).append(c); // 记录最后一个字符的计数和字符s = sb.toString();}return s;}
}
Python
class Solution:def countAndSay(self, n: int) -> str:s = "1"for _ in range(n - 1):i, new_s = 0, ""while i < len(s):count = 1while i + 1 < len(s) and s[i] == s[i + 1]:i += 1count += 1new_s += str(count) + s[i]i += 1s = new_sreturn s