|
67 | 67 |
|
68 | 68 | <!-- 这里可写通用的实现逻辑 --> |
69 | 69 |
|
| 70 | +**方法一：枚举** |
| 71 | + |
| 72 | +我们可以枚举矩阵的每个位置 $(i, j)$，判断是否能以该位置为起点，从左到右或从右到左放置单词 `word`，或者从上到下或从下到上放置单词 `word`。 |
| 73 | + |
| 74 | +该位置能作为起点需要满足以下条件： |
| 75 | + |
| 76 | +1. 如果要从从左到右放置单词 `word`，那么该位置必须是左边界，或者该位置左边的格子 `board[i][j - 1]` 是 `'#'`； |
| 77 | +2. 如果要从从右到左放置单词 `word`，那么该位置必须是右边界，或者该位置右边的格子 `board[i][j + 1]` 是 `'#'`； |
| 78 | +3. 如果要从从上到下放置单词 `word`，那么该位置必须是上边界，或者该位置上边的格子 `board[i - 1][j]` 是 `'#'`； |
| 79 | +4. 如果要从从下到上放置单词 `word`，那么该位置必须是下边界，或者该位置下边的格子 `board[i + 1][j]` 是 `'#'`。 |
| 80 | + |
| 81 | +在满足上述条件的情况下，我们可以从该位置开始，判断是否能放置单词 `word`。我们设计一个函数 $check(i, j, a, b)$，表示从位置 $(i, j)$ 开始，沿着方向 $(a, b)$ 放置单词 `word` 是否合法。如果合法，返回 `true`，否则返回 `false`。 |
| 82 | + |
| 83 | +函数 $check(i, j, a, b)$ 的实现如下： |
| 84 | + |
| 85 | +我们先获取当前方向的另一个边界位置 $(x, y)$，即 $(x, y) = (i + a \times k, j + b \times k)$，其中 $k$ 为单词 `word` 的长度。如果 $(x, y)$ 在矩阵内，并且 $(x, y)$ 的格子不是 `'#'`，则说明当前方向的另一个边界位置不是 `'#'`，因此不能放置单词 `word`，返回 `false`。 |
| 86 | + |
| 87 | +否则，我们从位置 $(i, j)$ 开始，沿着方向 $(a, b)$ 遍历单词 `word`，如果遇到格子 `board[i][j]` 不是空格或者不是单词 `word` 的当前字符，说明不能放置单词 `word`，返回 `false`。如果遍历完单词 `word`，说明能放置单词 `word`，返回 `true`。 |
| 88 | + |
| 89 | +时间复杂度 $O(m \times n)$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。 |
| 90 | + |
70 | 91 | <!-- tabs:start --> |
71 | 92 |
|
72 | 93 | ### **Python3** |
73 | 94 |
|
74 | 95 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
75 | 96 |
|
76 | 97 | ```python |
77 | | - |
| 98 | +class Solution: |
| 99 | + def placeWordInCrossword(self, board: List[List[str]], word: str) -> bool: |
| 100 | + def check(i, j, a, b): |
| 101 | + x, y = i + a * k, j + b * k |
| 102 | + if 0 <= x < m and 0 <= y < n and board[x][y] != '#': |
| 103 | + return False |
| 104 | + for c in word: |
| 105 | + if ( |
| 106 | + i < 0 |
| 107 | + or i >= m |
| 108 | + or j < 0 |
| 109 | + or j >= n |
| 110 | + or (board[i][j] != ' ' and board[i][j] != c) |
| 111 | + ): |
| 112 | + return False |
| 113 | + i, j = i + a, j + b |
| 114 | + return True |
| 115 | + |
| 116 | + m, n = len(board), len(board[0]) |
| 117 | + k = len(word) |
| 118 | + for i in range(m): |
| 119 | + for j in range(n): |
| 120 | + left_to_right = (j == 0 or board[i][j - 1] == '#') and check(i, j, 0, 1) |
| 121 | + right_to_left = (j == n - 1 or board[i][j + 1] == '#') and check( |
| 122 | + i, j, 0, -1 |
| 123 | + ) |
| 124 | + up_to_down = (i == 0 or board[i - 1][j] == '#') and check(i, j, 1, 0) |
| 125 | + down_to_up = (i == m - 1 or board[i + 1][j] == '#') and check( |
| 126 | + i, j, -1, 0 |
| 127 | + ) |
| 128 | + if left_to_right or right_to_left or up_to_down or down_to_up: |
| 129 | + return True |
| 130 | + return False |
78 | 131 | ``` |
79 | 132 |
|
80 | 133 | ### **Java** |
81 | 134 |
|
82 | 135 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
83 | 136 |
|
84 | 137 | ```java |
| 138 | +class Solution { |
| 139 | + private int m; |
| 140 | + private int n; |
| 141 | + private char[][] board; |
| 142 | + private String word; |
| 143 | + private int k; |
| 144 | + |
| 145 | + public boolean placeWordInCrossword(char[][] board, String word) { |
| 146 | + m = board.length; |
| 147 | + n = board[0].length; |
| 148 | + this.board = board; |
| 149 | + this.word = word; |
| 150 | + k = word.length(); |
| 151 | + for (int i = 0; i < m; ++i) { |
| 152 | + for (int j = 0; j < n; ++j) { |
| 153 | + boolean leftToRight = (j == 0 || board[i][j - 1] == '#') && check(i, j, 0, 1); |
| 154 | + boolean rightToLeft = (j == n - 1 || board[i][j + 1] == '#') && check(i, j, 0, -1); |
| 155 | + boolean upToDown = (i == 0 || board[i - 1][j] == '#') && check(i, j, 1, 0); |
| 156 | + boolean downToUp = (i == m - 1 || board[i + 1][j] == '#') && check(i, j, -1, 0); |
| 157 | + if (leftToRight || rightToLeft || upToDown || downToUp) { |
| 158 | + return true; |
| 159 | + } |
| 160 | + } |
| 161 | + } |
| 162 | + return false; |
| 163 | + } |
| 164 | + |
| 165 | + private boolean check(int i, int j, int a, int b) { |
| 166 | + int x = i + a * k, y = j + b * k; |
| 167 | + if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#') { |
| 168 | + return false; |
| 169 | + } |
| 170 | + for (int p = 0; p < k; ++p) { |
| 171 | + if (i < 0 || i >= m || j < 0 || j >= n || (board[i][j] != ' ' && board[i][j] != word.charAt(p))) { |
| 172 | + return false; |
| 173 | + } |
| 174 | + i += a; |
| 175 | + j += b; |
| 176 | + } |
| 177 | + return true; |
| 178 | + } |
| 179 | +} |
| 180 | +``` |
| 181 | + |
| 182 | +### **C++** |
| 183 | + |
| 184 | +```cpp |
| 185 | +class Solution { |
| 186 | +public: |
| 187 | + bool placeWordInCrossword(vector<vector<char>>& board, string word) { |
| 188 | + int m = board.size(), n = board[0].size(); |
| 189 | + int k = word.size(); |
| 190 | + auto check = [&](int i, int j, int a, int b) { |
| 191 | + int x = i + a * k, y = j + b * k; |
| 192 | + if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#') { |
| 193 | + return false; |
| 194 | + } |
| 195 | + for (char& c : word) { |
| 196 | + if (i < 0 || i >= m || j < 0 || j >= n || (board[i][j] != ' ' && board[i][j] != c)) { |
| 197 | + return false; |
| 198 | + } |
| 199 | + i += a; |
| 200 | + j += b; |
| 201 | + } |
| 202 | + return true; |
| 203 | + }; |
| 204 | + for (int i = 0; i < m; ++i) { |
| 205 | + for (int j = 0; j < n; ++j) { |
| 206 | + bool leftToRight = (j == 0 || board[i][j - 1] == '#') && check(i, j, 0, 1); |
| 207 | + bool rightToLeft = (j == n - 1 || board[i][j + 1] == '#') && check(i, j, 0, -1); |
| 208 | + bool upToDown = (i == 0 || board[i - 1][j] == '#') && check(i, j, 1, 0); |
| 209 | + bool downToUp = (i == m - 1 || board[i + 1][j] == '#') && check(i, j, -1, 0); |
| 210 | + if (leftToRight || rightToLeft || upToDown || downToUp) { |
| 211 | + return true; |
| 212 | + } |
| 213 | + } |
| 214 | + } |
| 215 | + return false; |
| 216 | + } |
| 217 | +}; |
| 218 | +``` |
85 | 219 |
|
| 220 | +### **Go** |
| 221 | +
|
| 222 | +```go |
| 223 | +func placeWordInCrossword(board [][]byte, word string) bool { |
| 224 | + m, n := len(board), len(board[0]) |
| 225 | + k := len(word) |
| 226 | + check := func(i, j, a, b int) bool { |
| 227 | + x, y := i+a*k, j+b*k |
| 228 | + if x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#' { |
| 229 | + return false |
| 230 | + } |
| 231 | + for _, c := range word { |
| 232 | + if i < 0 || i >= m || j < 0 || j >= n || (board[i][j] != ' ' && board[i][j] != byte(c)) { |
| 233 | + return false |
| 234 | + } |
| 235 | + i, j = i+a, j+b |
| 236 | + } |
| 237 | + return true |
| 238 | + } |
| 239 | + for i := range board { |
| 240 | + for j := range board[i] { |
| 241 | + leftToRight := (j == 0 || board[i][j-1] == '#') && check(i, j, 0, 1) |
| 242 | + rightToLeft := (j == n-1 || board[i][j+1] == '#') && check(i, j, 0, -1) |
| 243 | + upToDown := (i == 0 || board[i-1][j] == '#') && check(i, j, 1, 0) |
| 244 | + downToUp := (i == m-1 || board[i+1][j] == '#') && check(i, j, -1, 0) |
| 245 | + if leftToRight || rightToLeft || upToDown || downToUp { |
| 246 | + return true |
| 247 | + } |
| 248 | + } |
| 249 | + } |
| 250 | + return false |
| 251 | +} |
86 | 252 | ``` |
87 | 253 |
|
88 | 254 | ### **...** |
|
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