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Commit f108f8b

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feat: add solutions to lc problem: No.0886.Possible Bipartition
1 parent c0f6a38 commit f108f8b

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‎index.html‎

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<script>
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window.$docsify = {
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name: 'leetcode',
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repo: 'doocs/leetcode',
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logo: '/images/doocs-leetcode.png',
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search: [
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'/', '/solution/', '/lcof/', '/lcof2/', '/lcci/', '/lcs/', '/lcp/', '/basic/'

‎solution/0800-0899/0886.Possible Bipartition/README.md‎

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<li>对于 <code>dislikes[i] == dislikes[j]</code> 不存在 <code>i != j</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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并查集。
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模板 1——朴素并查集:
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```python
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# 初始化,p存储每个点的父节点
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p = list(range(n))
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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# 路径压缩
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p[x] = find(p[x])
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return p[x]
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# 合并a和b所在的两个集合
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p[find(a)] = find(b)
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```
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模板 2——维护 size 的并查集:
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```python
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# 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
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p = list(range(n))
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size = [1] * n
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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# 路径压缩
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p[x] = find(p[x])
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return p[x]
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# 合并a和b所在的两个集合
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if find(a) != find(b):
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size[find(b)] += size[find(a)]
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p[find(a)] = find(b)
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```
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模板 3——维护到祖宗节点距离的并查集:
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x
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```python
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# 初始化,p存储每个点的父节点,d[x]存储x到p[x]的距离
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p = list(range(n))
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d = [0] * n
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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t = find(p[x])
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d[x] += d[p[x]]
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p[x] = t
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return p[x]
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# 合并a和b所在的两个集合
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p[find(a)] = find(b)
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d[find(a)] = distance
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```
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def possibleBipartition(self, n: int, dislikes: List[List[int]]) -> bool:
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p = list(range(n))
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def find(x):
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if p[x] != x:
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p[x] = find(p[x])
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return p[x]
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mp = collections.defaultdict(list)
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for i, j in dislikes:
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mp[i - 1].append(j - 1)
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mp[j - 1].append(i - 1)
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for i in range(n):
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dis = mp[i]
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for j in dis:
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if find(i) == find(j):
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return False
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p[find(j)] = find(dis[0])
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return True
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public:
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vector<int> p;
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bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
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p.resize(n);
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for (int i = 0; i < n; ++i) p[i] = i;
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unordered_map<int, vector<int>> mp;
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for (auto e : dislikes)
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{
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mp[e[0] - 1].push_back(e[1] - 1);
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mp[e[1] - 1].push_back(e[0] - 1);
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}
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for (int i = 0; i < n; ++i)
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{
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auto dis = mp[i];
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for (int j : dis)
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{
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if (find(i) == find(j)) return false;
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p[find(j)] = find(dis[0]);
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}
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}
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return true;
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}
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int find(int x) {
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if (p[x] != x) p[x] = find(p[x]);
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return p[x];
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}
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};
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> p;
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bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
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p.resize(n);
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for (int i = 0; i < n; ++i) p[i] = i;
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unordered_map<int, vector<int>> mp;
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for (auto e : dislikes)
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{
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mp[e[0] - 1].push_back(e[1] - 1);
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mp[e[1] - 1].push_back(e[0] - 1);
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}
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for (int i = 0; i < n; ++i)
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{
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auto dis = mp[i];
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for (int j : dis)
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{
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if (find(i) == find(j)) return false;
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p[find(j)] = find(dis[0]);
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}
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}
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return true;
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}
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int find(int x) {
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if (p[x] != x) p[x] = find(p[x]);
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return p[x];
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}
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};
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```
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### **Go**
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```go
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var p []int
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func possibleBipartition(n int, dislikes [][]int) bool {
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p = make([]int, n)
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for i := 0; i < n; i++ {
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p[i] = i
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}
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mp := make(map[int][]int)
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for _, e := range dislikes {
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mp[e[0]-1] = append(mp[e[0]-1], e[1]-1)
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mp[e[1]-1] = append(mp[e[1]-1], e[0]-1)
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}
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for i := 0; i < n; i++ {
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dis := mp[i]
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for _, j := range dis {
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if find(i) == find(j) {
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return false
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}
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p[find(j)] = find(dis[0])
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}
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}
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return true
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}
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func find(x int) int {
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if p[x] != x {
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p[x] = find(p[x])
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}
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return p[x]
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}
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```
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### **...**

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