14 并查集

Posted 2021-12-28 lijunjie9502

• 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）

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

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

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

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

• letcode 例题
  • https://leetcode.com/problems/number-of-islands/

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

   

• https://leetcode.com/problems/friend-circles/submissions/

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