|
75 | 75 |
|
76 | 76 | ## 解法 |
77 | 77 |
|
78 | | -### 方法一 |
| 78 | +### 方法一:哈希表 + 枚举 |
| 79 | + |
| 80 | +我们可以使用哈希表来统计每个大于 10 的素数出现的次数。 |
| 81 | + |
| 82 | +对于每个单元格,我们可以从它出发,沿着 8 个方向之一生成数字,然后判断生成的数字是否是大于 10ドル$ 的素数,如果是的话,就将它加入到哈希表中。 |
| 83 | + |
| 84 | +最后,我们遍历哈希表,找到出现频率最高的素数,如果有多个出现频率最高的素数,那么返回其中最大的那个。 |
| 85 | + |
| 86 | +时间复杂度 $O(m \times n \times \max(m, n) \times {10}^{\frac{\max(m, n)}{2}}),ドル空间复杂度 $O(m \times n \times \max(m, n))$。其中 $m$ 和 $n$ 分别是 `mat` 的行数和列数。 |
79 | 87 |
|
80 | 88 | <!-- tabs:start --> |
81 | 89 |
|
82 | 90 | ```python |
| 91 | +class Solution: |
| 92 | + def mostFrequentPrime(self, mat: List[List[int]]) -> int: |
| 93 | + def is_prime(x: int) -> int: |
| 94 | + return all(x % i != 0 for i in range(2, isqrt(x) + 1)) |
83 | 95 |
|
| 96 | + m, n = len(mat), len(mat[0]) |
| 97 | + cnt = Counter() |
| 98 | + for i in range(m): |
| 99 | + for j in range(n): |
| 100 | + for a in range(-1, 2): |
| 101 | + for b in range(-1, 2): |
| 102 | + if a == 0 and b == 0: |
| 103 | + continue |
| 104 | + x, y, v = i + a, j + b, mat[i][j] |
| 105 | + while 0 <= x < m and 0 <= y < n: |
| 106 | + v = v * 10 + mat[x][y] |
| 107 | + if is_prime(v): |
| 108 | + cnt[v] += 1 |
| 109 | + x, y = x + a, y + b |
| 110 | + ans, mx = -1, 0 |
| 111 | + for v, x in cnt.items(): |
| 112 | + if mx < x: |
| 113 | + mx = x |
| 114 | + ans = v |
| 115 | + elif mx == x: |
| 116 | + ans = max(ans, v) |
| 117 | + return ans |
84 | 118 | ``` |
85 | 119 |
|
86 | 120 | ```java |
| 121 | +class Solution { |
| 122 | + public int mostFrequentPrime(int[][] mat) { |
| 123 | + int m = mat.length, n = mat[0].length; |
| 124 | + Map<Integer, Integer> cnt = new HashMap<>(); |
| 125 | + for (int i = 0; i < m; ++i) { |
| 126 | + for (int j = 0; j < n; ++j) { |
| 127 | + for (int a = -1; a <= 1; ++a) { |
| 128 | + for (int b = -1; b <= 1; ++b) { |
| 129 | + if (a == 0 && b == 0) { |
| 130 | + continue; |
| 131 | + } |
| 132 | + int x = i + a, y = j + b, v = mat[i][j]; |
| 133 | + while (x >= 0 && x < m && y >= 0 && y < n) { |
| 134 | + v = v * 10 + mat[x][y]; |
| 135 | + if (isPrime(v)) { |
| 136 | + cnt.merge(v, 1, Integer::sum); |
| 137 | + } |
| 138 | + x += a; |
| 139 | + y += b; |
| 140 | + } |
| 141 | + } |
| 142 | + } |
| 143 | + } |
| 144 | + } |
| 145 | + int ans = -1, mx = 0; |
| 146 | + for (var e : cnt.entrySet()) { |
| 147 | + int v = e.getKey(), x = e.getValue(); |
| 148 | + if (mx < x || (mx == x && ans < v)) { |
| 149 | + mx = x; |
| 150 | + ans = v; |
| 151 | + } |
| 152 | + } |
| 153 | + return ans; |
| 154 | + } |
87 | 155 |
|
| 156 | + private boolean isPrime(int n) { |
| 157 | + for (int i = 2; i <= n / i; ++i) { |
| 158 | + if (n % i == 0) { |
| 159 | + return false; |
| 160 | + } |
| 161 | + } |
| 162 | + return true; |
| 163 | + } |
| 164 | +} |
88 | 165 | ``` |
89 | 166 |
|
90 | 167 | ```cpp |
| 168 | +class Solution { |
| 169 | +public: |
| 170 | + int mostFrequentPrime(vector<vector<int>>& mat) { |
| 171 | + int m = mat.size(), n = mat[0].size(); |
| 172 | + unordered_map<int, int> cnt; |
| 173 | + for (int i = 0; i < m; ++i) { |
| 174 | + for (int j = 0; j < n; ++j) { |
| 175 | + for (int a = -1; a <= 1; ++a) { |
| 176 | + for (int b = -1; b <= 1; ++b) { |
| 177 | + if (a == 0 && b == 0) { |
| 178 | + continue; |
| 179 | + } |
| 180 | + int x = i + a, y = j + b, v = mat[i][j]; |
| 181 | + while (x >= 0 && x < m && y >= 0 && y < n) { |
| 182 | + v = v * 10 + mat[x][y]; |
| 183 | + if (isPrime(v)) { |
| 184 | + cnt[v]++; |
| 185 | + } |
| 186 | + x += a; |
| 187 | + y += b; |
| 188 | + } |
| 189 | + } |
| 190 | + } |
| 191 | + } |
| 192 | + } |
| 193 | + int ans = -1, mx = 0; |
| 194 | + for (auto& [v, x] : cnt) { |
| 195 | + if (mx < x || (mx == x && ans < v)) { |
| 196 | + mx = x; |
| 197 | + ans = v; |
| 198 | + } |
| 199 | + } |
| 200 | + return ans; |
| 201 | + } |
91 | 202 |
|
| 203 | +private: |
| 204 | + bool isPrime(int n) { |
| 205 | + for (int i = 2; i <= n / i; ++i) { |
| 206 | + if (n % i == 0) { |
| 207 | + return false; |
| 208 | + } |
| 209 | + } |
| 210 | + return true; |
| 211 | + } |
| 212 | +}; |
92 | 213 | ``` |
93 | 214 | |
94 | 215 | ```go |
| 216 | +func mostFrequentPrime(mat [][]int) int { |
| 217 | + m, n := len(mat), len(mat[0]) |
| 218 | + cnt := make(map[int]int) |
| 219 | + for i := 0; i < m; i++ { |
| 220 | + for j := 0; j < n; j++ { |
| 221 | + for a := -1; a <= 1; a++ { |
| 222 | + for b := -1; b <= 1; b++ { |
| 223 | + if a == 0 && b == 0 { |
| 224 | + continue |
| 225 | + } |
| 226 | + x, y, v := i+a, j+b, mat[i][j] |
| 227 | + for x >= 0 && x < m && y >= 0 && y < n { |
| 228 | + v = v*10 + mat[x][y] |
| 229 | + if isPrime(v) { |
| 230 | + cnt[v]++ |
| 231 | + } |
| 232 | + x += a |
| 233 | + y += b |
| 234 | + } |
| 235 | + } |
| 236 | + } |
| 237 | + } |
| 238 | + } |
| 239 | + ans, mx := -1, 0 |
| 240 | + for v, x := range cnt { |
| 241 | + if mx < x || (mx == x && ans < v) { |
| 242 | + mx = x |
| 243 | + ans = v |
| 244 | + } |
| 245 | + } |
| 246 | + return ans |
| 247 | +} |
| 248 | + |
| 249 | +func isPrime(n int) bool { |
| 250 | + for i := 2; i <= n/i; i++ { |
| 251 | + if n%i == 0 { |
| 252 | + return false |
| 253 | + } |
| 254 | + } |
| 255 | + return true |
| 256 | +} |
| 257 | +``` |
| 258 | + |
| 259 | +```ts |
| 260 | +function mostFrequentPrime(mat: number[][]): number { |
| 261 | + const m: number = mat.length; |
| 262 | + const n: number = mat[0].length; |
| 263 | + const cnt: Map<number, number> = new Map(); |
| 264 | + const isPrime = (x: number): boolean => { |
| 265 | + for (let i = 2; i <= x / i; ++i) { |
| 266 | + if (x % i === 0) { |
| 267 | + return false; |
| 268 | + } |
| 269 | + } |
| 270 | + return true; |
| 271 | + }; |
| 272 | + |
| 273 | + for (let i = 0; i < m; ++i) { |
| 274 | + for (let j = 0; j < n; ++j) { |
| 275 | + for (let a = -1; a <= 1; ++a) { |
| 276 | + for (let b = -1; b <= 1; ++b) { |
| 277 | + if (a === 0 && b === 0) { |
| 278 | + continue; |
| 279 | + } |
| 280 | + let [x, y, v] = [i + a, j + b, mat[i][j]]; |
| 281 | + while (x >= 0 && x < m && y >= 0 && y < n) { |
| 282 | + v = v * 10 + mat[x][y]; |
| 283 | + if (isPrime(v)) { |
| 284 | + cnt.set(v, (cnt.get(v) || 0) + 1); |
| 285 | + } |
| 286 | + x += a; |
| 287 | + y += b; |
| 288 | + } |
| 289 | + } |
| 290 | + } |
| 291 | + } |
| 292 | + } |
95 | 293 |
|
| 294 | + let [ans, mx] = [-1, 0]; |
| 295 | + cnt.forEach((x, v) => { |
| 296 | + if (mx < x || (mx === x && ans < v)) { |
| 297 | + mx = x; |
| 298 | + ans = v; |
| 299 | + } |
| 300 | + }); |
| 301 | + return ans; |
| 302 | +} |
96 | 303 | ``` |
97 | 304 |
|
98 | 305 | <!-- tabs:end --> |
|
0 commit comments