标签：src add function 例题 lse python 基本概念 lap 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
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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 |
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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 |
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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 |
标签：src add function 例题 lse python 基本概念 lap 伪代码
原文地址：https://www.cnblogs.com/lijunjie9502/p/10989224.html
