- 基本概念
- 并查集是一种树型的数据结构，用于处理一些不交集的合并及查询问题
- 示意图
- 常用方法
- Find：确定元素属于哪一个子集。可以用来确定两个元素是否属于同一子集
- Union： 将两个子集合并成同一个集合
- 示意图
- 代码实现（伪代码）
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function MakeSet(x) x.parent := x function Find(x) if x.parent == x return x else return Find(x.parent) function Union(x, y) xRoot := Find(x) yRoot := Find(y) xRoot.parent := yRoot |
- 优化
- rank
- 示意图
- 伪代码
- 示意图
- rank
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function MakeSet(x) x.parent := x x.rank := 0 function Union(x, y) xRoot := Find(x) yRoot := Find(y) if xRoot == yRoot return if xRoot.rank < yRoot.rank xRoot.parent := yRoot else if xRoot.rank > yRoot.rank yRoot.parent := xRoot else yRoot.parent := xRoot xRoot.rank := xRoot.rank + 1 |
- 路径压缩
- 示意图
- 伪代码
- 示意图
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function MakeSet(x) x.parent := x function Find(x) root = x while root != root.parent root := root.parent while x != x.parent tmp = x.parent x.parent = root x = tmp return root |
- Python 实现 -- 包含两种优化
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class UnionUF(object): def __init__(self, n): self.parent = [-1] * n self.rank = [0] * n for i in range(n): self.parent[i] = i def find(self, i): root = i while root != self.parent[root]: root = self.parent[root] # 路径压缩 while i != self.parent[i]: i, self.parent[i] = self.parent[i], root return root def union(self, x, y): rootx = self.find(x) rooty = self.find(y) if rootx == rooty: return 0# 两者属于同一集合,没有发生合并操作 if self.rank[rootx] > self.rank[rooty]: self.parent[rooty] = rootx elif self.rank[rootx] < self.rank[rooty]: self.parent[rootx] = rooty else: self.parent[rootx] = rooty self.rank[rooty] += 1 return 1# 发生合并操作 |
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class UnionUF(object): def __init__(self, n): self.parent = [-1] * n self.rank = [0] * n for i in range(n): self.parent[i] = i def find(self, i): root = i while root != self.parent[root]: root = self.parent[root] # 路径压缩 while i != self.parent[i]: i, self.parent[i] = self.parent[i], root return root def union(self, x, y): rootx = self.find(x) rooty = self.find(y) if rootx == rooty: return 0# 两者属于同一集合,没有发生合并操作 if self.rank[rootx] > self.rank[rooty]: self.parent[rooty] = rootx elif self.rank[rootx] < self.rank[rooty]: self.parent[rootx] = rooty else: self.parent[rootx] = rooty self.rank[rooty] += 1 return 1# 发生合并操作 class Solution: def numIslands(self, grid: List[List[str]]) -> int: if not grid: return 0 m, n = len(grid), len(grid[0]) union = UnionUF(m*n) res = 0 dx, dy = [-1, 1, 0, 0], [0, 0, -1, 1] for i in range(m): for j in range(n): if grid[i][j] == "1": res += 1 for count in range(len(dx)): ni, nj = i + dx[count], j + dy[count] if 0 <= ni < m and 0 <= nj < n and grid[ni][nj] == "1": res -= union.union(i*n + j, ni*n + nj) return res |
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class UnionUF(object): def __init__(self, n): self.parent = [-1] * n self.rank = [0] * n for i in range(n): self.parent[i] = i def find(self, i): root = i while root != self.parent[root]: root = self.parent[root] # 路径压缩 while i != self.parent[i]: i, self.parent[i] = self.parent[i], root return root def union(self, x, y): rootx = self.find(x) rooty = self.find(y) if rootx == rooty: return 0# 两者属于同一集合,没有发生合并操作 if self.rank[rootx] > self.rank[rooty]: self.parent[rooty] = rootx elif self.rank[rootx] < self.rank[rooty]: self.parent[rootx] = rooty else: self.parent[rootx] = rooty self.rank[rooty] += 1 return 1# 发生合并操作 class Solution: def findCircleNum(self, M: List[List[int]]) -> int: if not M: return 0 m = len(M) union = UnionUF(m) res = 0 for i in range(m): res += 1 for count in range(m): if i == count: continue if M[i][count] == 1: res -= union.union(i, count) return res |
原文地址：https://www.cnblogs.com/lijunjie9502/p/10989224.html