|
52 | 52 | <li><code>2 <= nums[i] <= 10<sup>5</sup></code></li> |
53 | 53 | </ul> |
54 | 54 |
|
55 | | - |
56 | 55 | ## 解法 |
57 | 56 |
|
58 | 57 | <!-- 这里可写通用的实现逻辑 --> |
59 | 58 |
|
| 59 | +并查集。 |
| 60 | + |
| 61 | +并查集模板: |
| 62 | + |
| 63 | +模板 1——朴素并查集: |
| 64 | + |
| 65 | +```python |
| 66 | +# 初始化,p存储每个点的父节点 |
| 67 | +p = list(range(n)) |
| 68 | + |
| 69 | +# 返回x的祖宗节点 |
| 70 | +def find(x): |
| 71 | + if p[x] != x: |
| 72 | + # 路径压缩 |
| 73 | + p[x] = find(p[x]) |
| 74 | + return p[x] |
| 75 | + |
| 76 | +# 合并a和b所在的两个集合 |
| 77 | +p[find(a)] = find(b) |
| 78 | +``` |
| 79 | + |
| 80 | +模板 2——维护 size 的并查集: |
| 81 | + |
| 82 | +```python |
| 83 | +# 初始化,p存储每个点的父节点,size只有当节点是祖宗节点时才有意义,表示祖宗节点所在集合中,点的数量 |
| 84 | +p = list(range(n)) |
| 85 | +size = [1] * n |
| 86 | + |
| 87 | +# 返回x的祖宗节点 |
| 88 | +def find(x): |
| 89 | + if p[x] != x: |
| 90 | + # 路径压缩 |
| 91 | + p[x] = find(p[x]) |
| 92 | + return p[x] |
| 93 | + |
| 94 | +# 合并a和b所在的两个集合 |
| 95 | +if find(a) != find(b): |
| 96 | + size[find(b)] += size[find(a)] |
| 97 | + p[find(a)] = find(b) |
| 98 | +``` |
| 99 | + |
| 100 | +模板 3——维护到祖宗节点距离的并查集: |
| 101 | + |
| 102 | +```python |
| 103 | +# 初始化,p存储每个点的父节点,d[x]存储x到p[x]的距离 |
| 104 | +p = list(range(n)) |
| 105 | +d = [0] * n |
| 106 | + |
| 107 | +# 返回x的祖宗节点 |
| 108 | +def find(x): |
| 109 | + if p[x] != x: |
| 110 | + t = find(p[x]) |
| 111 | + d[x] += d[p[x]] |
| 112 | + p[x] = t |
| 113 | + return p[x] |
| 114 | + |
| 115 | +# 合并a和b所在的两个集合 |
| 116 | +p[find(a)] = find(b) |
| 117 | +d[find(a)] = distance |
| 118 | +``` |
| 119 | + |
| 120 | +对于本题,最大公因数大于 1 的两个数,可以进行交换,因此,只要一个集合中所有数都存在相同公因数,那么这个集合中任意数都能进行两两交换,因此可以用并查集,把同个集合中的所有数字进行合并。 |
| 121 | + |
| 122 | +> 在这道题中,可以先预处理每个数的质因数,数字与质因数归属同一个集合。 |
| 123 | + |
| 124 | +合并之后,将原数组复制一份,并进行升序排列,得到新数组 s。然后遍历原数组,若原数组对应元素与新数组对应元素不相同,并且两个元素也不在同一个集合中,说明不满足条件,直接返回 false,否则遍历结束返回 true。 |
| 125 | + |
60 | 126 | <!-- tabs:start --> |
61 | 127 |
|
62 | 128 | ### **Python3** |
63 | 129 |
|
64 | 130 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
65 | 131 |
|
66 | 132 | ```python |
67 | | - |
| 133 | +class Solution: |
| 134 | + def gcdSort(self, nums: List[int]) -> bool: |
| 135 | + n = 10 ** 5 + 10 |
| 136 | + p = list(range(n)) |
| 137 | + f = collections.defaultdict(list) |
| 138 | + mx = max(nums) |
| 139 | + for i in range(2, mx + 1): |
| 140 | + if f[i]: |
| 141 | + continue |
| 142 | + for j in range(i, mx + 1, i): |
| 143 | + f[j].append(i) |
| 144 | + |
| 145 | + def find(x): |
| 146 | + if p[x] != x: |
| 147 | + p[x] = find(p[x]) |
| 148 | + return p[x] |
| 149 | + |
| 150 | + for i in nums: |
| 151 | + for j in f[i]: |
| 152 | + p[find(i)] = find(j) |
| 153 | + |
| 154 | + s = sorted(nums) |
| 155 | + for i, num in enumerate(nums): |
| 156 | + if s[i] != num and find(num) != find(s[i]): |
| 157 | + return False |
| 158 | + return True |
68 | 159 | ``` |
69 | 160 |
|
70 | 161 | ### **Java** |
71 | 162 |
|
72 | 163 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
73 | 164 |
|
74 | 165 | ```java |
| 166 | +class Solution { |
| 167 | + private int[] p; |
| 168 | + |
| 169 | + public boolean gcdSort(int[] nums) { |
| 170 | + int n = 100010; |
| 171 | + p = new int[n]; |
| 172 | + Map<Integer, List<Integer>> f = new HashMap<>(); |
| 173 | + for (int i = 0; i < n; ++i) { |
| 174 | + p[i] = i; |
| 175 | + } |
| 176 | + int mx = 0; |
| 177 | + for (int num : nums) { |
| 178 | + mx = Math.max(mx, num); |
| 179 | + } |
| 180 | + for (int i = 2; i <= mx; ++i) { |
| 181 | + if (f.containsKey(i)) { |
| 182 | + continue; |
| 183 | + } |
| 184 | + for (int j = i; j <= mx; j += i) { |
| 185 | + f.putIfAbsent(j, new ArrayList<>()); |
| 186 | + f.get(j).add(i); |
| 187 | + } |
| 188 | + } |
| 189 | + for (int i : nums) { |
| 190 | + for (int j : f.get(i)) { |
| 191 | + p[find(i)] = find(j); |
| 192 | + } |
| 193 | + } |
| 194 | + int[] s = new int[nums.length]; |
| 195 | + System.arraycopy(nums, 0, s, 0, nums.length); |
| 196 | + Arrays.sort(s); |
| 197 | + for (int i = 0; i < nums.length; ++i) { |
| 198 | + if (s[i] != nums[i] && find(nums[i]) != find(s[i])) { |
| 199 | + return false; |
| 200 | + } |
| 201 | + } |
| 202 | + return true; |
| 203 | + } |
| 204 | + |
| 205 | + int find(int x) { |
| 206 | + if (p[x] != x) { |
| 207 | + p[x] = find(p[x]); |
| 208 | + } |
| 209 | + return p[x]; |
| 210 | + } |
| 211 | +} |
| 212 | +``` |
| 213 | + |
| 214 | +### **C++** |
| 215 | + |
| 216 | +```cpp |
| 217 | +class Solution { |
| 218 | +public: |
| 219 | + vector<int> p; |
| 220 | + |
| 221 | + bool gcdSort(vector<int>& nums) { |
| 222 | + int n = 100010; |
| 223 | + p.resize(n); |
| 224 | + for (int i = 0; i < n; ++i) p[i] = i; |
| 225 | + int mx = 0; |
| 226 | + for (auto num : nums) mx = max(mx, num); |
| 227 | + unordered_map<int, vector<int>> f; |
| 228 | + for (int i = 2; i <= mx; ++i) |
| 229 | + { |
| 230 | + if (!f[i].empty()) continue; |
| 231 | + for (int j = i; j <= mx; j += i) f[j].push_back(i); |
| 232 | + } |
| 233 | + for (int i : nums) |
| 234 | + { |
| 235 | + for (int j : f[i]) p[find(i)] = find(j); |
| 236 | + } |
| 237 | + vector<int> s = nums; |
| 238 | + sort(s.begin(), s.end()); |
| 239 | + for (int i = 0; i < nums.size(); ++i) |
| 240 | + { |
| 241 | + if (s[i] != nums[i] && find(s[i]) != find(nums[i])) return false; |
| 242 | + } |
| 243 | + return true; |
| 244 | + } |
| 245 | + |
| 246 | + int find(int x) { |
| 247 | + if (p[x] != x) p[x] = find(p[x]); |
| 248 | + return p[x]; |
| 249 | + } |
| 250 | +}; |
| 251 | +``` |
75 | 252 |
|
| 253 | +### **Go** |
| 254 | + |
| 255 | +```go |
| 256 | +var p []int |
| 257 | + |
| 258 | +func gcdSort(nums []int) bool { |
| 259 | + n := 100010 |
| 260 | + p = make([]int, n) |
| 261 | + for i := 0; i < n; i++ { |
| 262 | + p[i] = i |
| 263 | + } |
| 264 | + mx := 0 |
| 265 | + for _, num := range nums { |
| 266 | + mx = max(mx, num) |
| 267 | + } |
| 268 | + f := make([][]int, mx+1) |
| 269 | + for i := 2; i <= mx; i++ { |
| 270 | + if len(f[i]) > 0 { |
| 271 | + continue |
| 272 | + } |
| 273 | + for j := i; j <= mx; j += i { |
| 274 | + f[j] = append(f[j], i) |
| 275 | + } |
| 276 | + } |
| 277 | + for _, i := range nums { |
| 278 | + for _, j := range f[i] { |
| 279 | + p[find(i)] = find(j) |
| 280 | + } |
| 281 | + } |
| 282 | + s := make([]int, len(nums)) |
| 283 | + for i, num := range nums { |
| 284 | + s[i] = num |
| 285 | + } |
| 286 | + sort.Ints(s) |
| 287 | + for i, num := range nums { |
| 288 | + if s[i] != num && find(s[i]) != find(num) { |
| 289 | + return false |
| 290 | + } |
| 291 | + } |
| 292 | + return true |
| 293 | +} |
| 294 | + |
| 295 | +func find(x int) int { |
| 296 | + if p[x] != x { |
| 297 | + p[x] = find(p[x]) |
| 298 | + } |
| 299 | + return p[x] |
| 300 | +} |
| 301 | + |
| 302 | +func max(a, b int) int { |
| 303 | + if a > b { |
| 304 | + return a |
| 305 | + } |
| 306 | + return b |
| 307 | +} |
76 | 308 | ``` |
77 | 309 |
|
78 | 310 | ### **...** |
|
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