解数独，单词搜索，被围绕的区域。每题做详细思路梳理，配套Python&Java双语代码， 2024.03.07 可通过leetcode所有测试用例。
目录
37. 解数独
解题思路
完整代码
Python
Java
79. 单词搜索
解题思路
完整代码
Python
Java
130. 被围绕的区域
解题思路
完整代码
Python
Java
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：
解题思路
解决数独问题通常可以使用回溯算法，这是一种深度优先搜索的策略，用于尝试填充数独的每个空格，直到找到有效解为止。算法的基本思想是：
- 从数独的第一个空格开始。
- 尝试在当前空格中填入1到9之间的任何一个数字。
- 检查当前填入的数字是否满足数独的规则（即该数字在当前行、列、以及3x3的宫内没有重复）。
- 如果当前数字有效，则递归地继续填写下一个空格。如果遇到某个空格没有有效数字可以填入，则回溯到上一个空格，更换上一个空格的数字，再次尝试。
- 重复上述过程，直到所有的空格都被正确填满，解决数独问题。
完整代码
Python
class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
def isValid(row, col, ch):
for i in range(9):
if board[i][col] == ch or board[row][i] == ch: # 检查行和列
return False
if board[3 * (row // 3) + i // 3][3 * (col // 3) + i % 3] == ch: # 检查3x3的宫
return False
return True
def backtrack(board):
for i in range(9):
for j in range(9):
if board[i][j] != '.': # 跳过非空格
continue
for k in range(1, 10):
if isValid(i, j, str(k)):
board[i][j] = str(k) # 做选择
if backtrack(board): # 如果找到一种解决方案
return True
board[i][j] = '.' # 撤销选择
return False # 试遍了1-9都不行，返回False
return True # 遍历完没有返回False，说明找到了解决方案
backtrack(board)
Java
class Solution {
public void solveSudoku(char[][] board) {
solve(board);
}
private boolean solve(char[][] board) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] != '.') continue;
for (char c = '1'; c <= '9'; c++) { // 尝试1到9
if (isValid(board, i, j, c)) {
board[i][j] = c; // 做选择
if (solve(board)) return true; // 如果找到一种解决方案
board[i][j] = '.'; // 撤销选择
}
}
return false; // 试遍1-9都不行，返回false
}
}
return true; // 找到解决方案
}
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;
}
}
79. 单词搜索
给定一个
m x n二维字符网格board和一个字符串单词word。如果word存在于网格中，返回true；否则，返回false。单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
示例 1：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" 输出：true示例 2：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" 输出：true示例 3：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" 输出：false提示：
m == board.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15board和word仅由大小写英文字母组成进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在
board更大的情况下可以更快解决问题？
解题思路
这个问题可以通过深度优先搜索(DFS)和回溯算法来解决。算法的基本思想是遍历二维网格的每一个格子，从每个格子开始，尝试所有可能的路径去匹配给定的单词。如果某条路径上的字符序列与单词匹配，则说明单词存在于网格中。为了确保同一个单元格内的字母不被重复使用，我们需要记录已经访问过的单元格位置，并在递归调用后撤销该记录（回溯）。搜索剪枝的技术可以用于优化解决方案，比如当当前路径上的字符序列与单词不匹配时，提前结束当前路径的搜索。
- 遍历二维网格的每个格子。
- 从当前格子开始，使用深度优先搜索遍历相邻的格子。
- 检查当前路径上的字符序列是否与单词匹配。
- 如果当前字符与单词中对应的字符不匹配，回溯。
- 如果路径上的所有字符都匹配且路径长度等于单词长度，返回
true。
- 如果遍历了所有可能的路径都没有找到匹配的单词，返回
false。
完整代码
Python
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
if not board:
return False
# 深度优先搜索函数
def dfs(board, i, j, word, index):
# 判断当前位置是否越界或者字符是否不匹配
if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or board[i][j] != word[index]:
return False
# 如果当前字符是单词的最后一个字符，则找到匹配的路径
if index == len(word) - 1:
return True
# 暂时标记当前单元格，防止重复访问
temp, board[i][j] = board[i][j], '#'
# 搜索四个方向
found = dfs(board, i + 1, j, word, index + 1) or \
dfs(board, i - 1, j, word, index + 1) or \
dfs(board, i, j + 1, word, index + 1) or \
dfs(board, i, j - 1, word, index + 1)
# 回溯，恢复单元格的原始值
board[i][j] = temp
return found
# 遍历整个网格
for i in range(len(board)):
for j in range(len(board[0])):
if dfs(board, i, j, word, 0): # 从每个单元格开始尝试搜索
return True
return False
Java
public class Solution {
public boolean exist(char[][] board, String word) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (dfs(board, i, j, word, 0)) {
return true;
}
}
}
return false;
}
private boolean dfs(char[][] board, int x, int y, String word, int index) {
if (index == word.length()) {
return true;
}
if (x < 0 || x >= board.length || y < 0 || y >= board[0].length || board[x][y] != word.charAt(index)) {
return false;
}
char temp = board[x][y];
board[x][y] = '/'; // 标记当前格子为已访问
boolean found = dfs(board, x + 1, y, word, index + 1) || dfs(board, x - 1, y, word, index + 1) ||
dfs(board, x, y + 1, word, index + 1) || dfs(board, x, y - 1, word, index + 1);
board[x][y] = temp; // 回溯，撤销标记
return found;
}
}
130. 被围绕的区域
给你一个
m x n的矩阵board，由若干字符'X'和'O'，找到所有被'X'围绕的区域，并将这些区域里所有的'O'用'X'填充。示例 1：
输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]] 输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]] 解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的'O'都不会被填充为'X'。 任何不在边界上，或不与边界上的'O'相连的'O'最终都会被填充为'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。示例 2：
输入：board = [["X"]] 输出：[["X"]]提示：
m == board.lengthn == board[i].length1 <= m, n <= 200board[i][j]为'X'或'O'
解题思路
从矩阵的边界开始，找到所有的'O'字符，并通过深度优先搜索（DFS）标记所有与这些边界'O'相连的内部'O'字符。这样，未被标记的'O'即为被'X'完全包围的区域，随后将它们填充为'X'。
-
边界探索：
- 首先，从矩阵的边界出发，找到所有位于边界上的 'O'。
- 对于每个边界上的 'O'，使用深度优先搜索（DFS）或广度优先搜索（BFS）来探索与之相连的所有 'O'（即在水平或垂直方向上相邻的 'O'）。
- 在探索过程中，将这些与边界上的 'O' 相连的 'O' 暂时标记为另一个字符（比如 'T'），以表示这些 'O' 不应该被填充为 'X'，因为它们不是完全被 'X' 围绕的。
-
填充内部 'O'：
- 遍历整个矩阵，将所有未标记的 'O'（即那些被 'X' 完全围绕的 'O'）替换为 'X'。
-
恢复标记：
- 再次遍历矩阵，将步骤1中暂时标记为 'T' 的所有单元格恢复为 'O'。
通过以上步骤，所有被 'X' 完全围绕的 'O' 都会被填充为 'X'，而与边界相连的 'O' 保持不变。
完整代码
Python
class Solution:
def solve(self, board: List[List[str]]) -> None:
if not board or not board[0]:
return
rows, cols = len(board), len(board[0])
def dfs(i, j):
if 0 <= i < rows and 0 <= j < cols and board[i][j] == 'O':
board[i][j] = 'M'
dfs(i + 1, j)
dfs(i - 1, j)
dfs(i, j + 1)
dfs(i, j - 1)
for i in range(rows):
dfs(i, 0)
dfs(i, cols - 1)
for j in range(cols):
dfs(0, j)
dfs(rows - 1, j)
for i in range(rows):
for j in range(cols):
if board[i][j] == 'O':
board[i][j] = 'X'
elif board[i][j] == 'M':
board[i][j] = 'O'
Java
public class Solution {
public void solve(char[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) return;
int rows = board.length, cols = board[0].length;
for (int i = 0; i < rows; i++) {
dfs(board, i, 0);
dfs(board, i, cols - 1);
}
for (int j = 0; j < cols; j++) {
dfs(board, 0, j);
dfs(board, rows - 1, j);
}
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (board[i][j] == 'O') board[i][j] = 'X';
else if (board[i][j] == 'M') board[i][j] = 'O';
}
}
}
private void dfs(char[][] board, int i, int j) {
int rows = board.length, cols = board[0].length;
if (i < 0 || i >= rows || j < 0 || j >= cols || board[i][j] != 'O') return;
board[i][j] = 'M';
dfs(board, i + 1, j);
dfs(board, i - 1, j);
dfs(board, i, j + 1);
dfs(board, i, j - 1);
}
}
