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59 | 59 |
|
60 | 60 | <!-- 这里可写通用的实现逻辑 -->
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61 | 61 |
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| 62 | +**方法一:预处理 + 枚举 + 栈 + 回溯** |
| 63 | + |
| 64 | +由于题目中 $nums[i]$ 的取值范围为 $[1, 50]$,因此我们可以预处理出每个数的所有互质数,记录在数组 $f$ 中,其中 $f[i]$ 表示 $i$ 的所有互质数。 |
| 65 | + |
| 66 | +接下来我们可以使用回溯的方法,从根节点开始遍历整棵树,对于每个节点 $i$,我们可以通过 $f$ 数组得到 $nums[i]$ 的所有互质数。然后我们枚举 $nums[i]$ 的所有互质数,找到已经出现过的且深度最大的祖先节点 $t$,即为 $i$ 的最近的互质祖先节点。这里我们可以用一个长度为 $51$ 的栈数组 $stks$ 来获取每个出现过的值 $v$ 的节点以及其深度。每个栈 $stks[v]$ 的栈顶元素就是最近的深度最大的祖先节点。 |
| 67 | + |
| 68 | +时间复杂度 $O(n)$,空间复杂度 $O(n)$。其中 $n$ 为节点个数。 |
| 69 | + |
62 | 70 | <!-- tabs:start -->
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63 | 71 |
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64 | 72 | ### **Python3**
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65 | 73 |
|
66 | 74 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
67 | 75 |
|
68 | 76 | ```python
|
69 |
| - |
| 77 | +class Solution: |
| 78 | + def getCoprimes(self, nums: List[int], edges: List[List[int]]) -> List[int]: |
| 79 | + def dfs(i, fa, depth): |
| 80 | + t = k = -1 |
| 81 | + for v in f[nums[i]]: |
| 82 | + stk = stks[v] |
| 83 | + if stk and stk[-1][1] > k: |
| 84 | + t, k = stk[-1] |
| 85 | + ans[i] = t |
| 86 | + for j in g[i]: |
| 87 | + if j != fa: |
| 88 | + stks[nums[i]].append((i, depth)) |
| 89 | + dfs(j, i, depth + 1) |
| 90 | + stks[nums[i]].pop() |
| 91 | + |
| 92 | + g = defaultdict(list) |
| 93 | + for u, v in edges: |
| 94 | + g[u].append(v) |
| 95 | + g[v].append(u) |
| 96 | + f = defaultdict(list) |
| 97 | + for i in range(1, 51): |
| 98 | + for j in range(1, 51): |
| 99 | + if gcd(i, j) == 1: |
| 100 | + f[i].append(j) |
| 101 | + stks = defaultdict(list) |
| 102 | + ans = [-1] * len(nums) |
| 103 | + dfs(0, -1, 0) |
| 104 | + return ans |
70 | 105 | ```
|
71 | 106 |
|
72 | 107 | ### **Java**
|
73 | 108 |
|
74 | 109 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
75 | 110 |
|
76 | 111 | ```java
|
| 112 | +class Solution { |
| 113 | + private List<Integer>[] g; |
| 114 | + private List<Integer>[] f; |
| 115 | + private Deque<int[]>[] stks; |
| 116 | + private int[] nums; |
| 117 | + private int[] ans; |
| 118 | + |
| 119 | + public int[] getCoprimes(int[] nums, int[][] edges) { |
| 120 | + int n = nums.length; |
| 121 | + g = new List[n]; |
| 122 | + Arrays.setAll(g, k -> new ArrayList<>()); |
| 123 | + for (var e : edges) { |
| 124 | + int u = e[0], v = e[1]; |
| 125 | + g[u].add(v); |
| 126 | + g[v].add(u); |
| 127 | + } |
| 128 | + f = new List[51]; |
| 129 | + stks = new Deque[51]; |
| 130 | + Arrays.setAll(f, k -> new ArrayList<>()); |
| 131 | + Arrays.setAll(stks, k -> new ArrayDeque<>()); |
| 132 | + for (int i = 1; i < 51; ++i) { |
| 133 | + for (int j = 1; j < 51; ++j) { |
| 134 | + if (gcd(i, j) == 1) { |
| 135 | + f[i].add(j); |
| 136 | + } |
| 137 | + } |
| 138 | + } |
| 139 | + this.nums = nums; |
| 140 | + ans = new int[n]; |
| 141 | + dfs(0, -1, 0); |
| 142 | + return ans; |
| 143 | + } |
| 144 | + |
| 145 | + private void dfs(int i, int fa, int depth) { |
| 146 | + int t = -1, k = -1; |
| 147 | + for (int v : f[nums[i]]) { |
| 148 | + var stk = stks[v]; |
| 149 | + if (!stk.isEmpty() && stk.peek()[1] > k) { |
| 150 | + t = stk.peek()[0]; |
| 151 | + k = stk.peek()[1]; |
| 152 | + } |
| 153 | + } |
| 154 | + ans[i] = t; |
| 155 | + for (int j : g[i]) { |
| 156 | + if (j != fa) { |
| 157 | + stks[nums[i]].push(new int[] {i, depth}); |
| 158 | + dfs(j, i, depth + 1); |
| 159 | + stks[nums[i]].pop(); |
| 160 | + } |
| 161 | + } |
| 162 | + } |
| 163 | + |
| 164 | + private int gcd(int a, int b) { |
| 165 | + return b == 0 ? a : gcd(b, a % b); |
| 166 | + } |
| 167 | +} |
| 168 | +``` |
| 169 | + |
| 170 | +### **C++** |
| 171 | + |
| 172 | +```cpp |
| 173 | +class Solution { |
| 174 | +public: |
| 175 | + vector<int> getCoprimes(vector<int>& nums, vector<vector<int>>& edges) { |
| 176 | + int n = nums.size(); |
| 177 | + vector<vector<int>> g(n); |
| 178 | + vector<vector<int>> f(51); |
| 179 | + vector<stack<pair<int, int>>> stks(51); |
| 180 | + for (auto& e : edges) { |
| 181 | + int u = e[0], v = e[1]; |
| 182 | + g[u].emplace_back(v); |
| 183 | + g[v].emplace_back(u); |
| 184 | + } |
| 185 | + for (int i = 1; i < 51; ++i) { |
| 186 | + for (int j = 1; j < 51; ++j) { |
| 187 | + if (__gcd(i, j) == 1) { |
| 188 | + f[i].emplace_back(j); |
| 189 | + } |
| 190 | + } |
| 191 | + } |
| 192 | + vector<int> ans(n); |
| 193 | + function<void(int, int, int)> dfs = [&](int i, int fa, int depth) { |
| 194 | + int t = -1, k = -1; |
| 195 | + for (int v : f[nums[i]]) { |
| 196 | + auto& stk = stks[v]; |
| 197 | + if (!stk.empty() && stk.top().second > k) { |
| 198 | + t = stk.top().first; |
| 199 | + k = stk.top().second; |
| 200 | + } |
| 201 | + } |
| 202 | + ans[i] = t; |
| 203 | + for (int j : g[i]) { |
| 204 | + if (j != fa) { |
| 205 | + stks[nums[i]].push({i, depth}); |
| 206 | + dfs(j, i, depth + 1); |
| 207 | + stks[nums[i]].pop(); |
| 208 | + } |
| 209 | + } |
| 210 | + }; |
| 211 | + dfs(0, -1, 0); |
| 212 | + return ans; |
| 213 | + } |
| 214 | +}; |
| 215 | +``` |
77 | 216 |
|
| 217 | +### **Go** |
| 218 | +
|
| 219 | +```go |
| 220 | +func getCoprimes(nums []int, edges [][]int) []int { |
| 221 | + n := len(nums) |
| 222 | + g := make([][]int, n) |
| 223 | + f := [51][]int{} |
| 224 | + type pair struct{ first, second int } |
| 225 | + stks := [51][]pair{} |
| 226 | + for _, e := range edges { |
| 227 | + u, v := e[0], e[1] |
| 228 | + g[u] = append(g[u], v) |
| 229 | + g[v] = append(g[v], u) |
| 230 | + } |
| 231 | + for i := 1; i < 51; i++ { |
| 232 | + for j := 1; j < 51; j++ { |
| 233 | + if gcd(i, j) == 1 { |
| 234 | + f[i] = append(f[i], j) |
| 235 | + } |
| 236 | + } |
| 237 | + } |
| 238 | + ans := make([]int, n) |
| 239 | + var dfs func(i, fa, depth int) |
| 240 | + dfs = func(i, fa, depth int) { |
| 241 | + t, k := -1, -1 |
| 242 | + for _, v := range f[nums[i]] { |
| 243 | + stk := stks[v] |
| 244 | + if len(stk) > 0 && stk[len(stk)-1].second > k { |
| 245 | + t, k = stk[len(stk)-1].first, stk[len(stk)-1].second |
| 246 | + } |
| 247 | + } |
| 248 | + ans[i] = t |
| 249 | + for _, j := range g[i] { |
| 250 | + if j != fa { |
| 251 | + stks[nums[i]] = append(stks[nums[i]], pair{i, depth}) |
| 252 | + dfs(j, i, depth+1) |
| 253 | + stks[nums[i]] = stks[nums[i]][:len(stks[nums[i]])-1] |
| 254 | + } |
| 255 | + } |
| 256 | + } |
| 257 | + dfs(0, -1, 0) |
| 258 | + return ans |
| 259 | +} |
| 260 | +
|
| 261 | +func gcd(a, b int) int { |
| 262 | + if b == 0 { |
| 263 | + return a |
| 264 | + } |
| 265 | + return gcd(b, a%b) |
| 266 | +} |
78 | 267 | ```
|
79 | 268 |
|
80 | 269 | ### **...**
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