力扣130题、990题（并查集算法）

基础知识：

https://leetcode-cn.com/problems/satisfiability-of-equality-equations/solution/shou-hui-tu-jie-shou-xie-unionfind-bing-cha-ji-bu-/

130、被围绕的区域

具体实现：

靠边的O都不可能被替换，与这些O连接的O也不会被替换，抽象出一个dummy节点，这些O是dummy节点的子节点

二维坐标映射到一维左边的常用技巧：

m是行数，n是列数，二维坐标（i，j）可以转换成i*n+y

代码：

class Solution:  
    class UnionFind:
        def __init__(self,totalNodes):
            self.parents = list(range(totalNodes))
        def union(self, index1, index2):
            root1 = self.find(index1)
            root2 = self.find(index2)
            if root2!=root1:
                self.parents[root2] = root1
        def find(self,node):
            while self.parents[node] != node:
                self.parents[node] = self.parents[self.parents[node]]
                node = self.parents[node]
            return node
            #当前节点的父节点 指向父节点的父节点.
            #保证一个连通区域最终的parents只有一个.
        def isConnected(self,node1,node2):
            return self.find(node1) == self.find(node2)
 
    def solve(self, board: List[List[str]]) -> None:
        if len(board)== 0 :
            return 
        m = len(board)
        n = len(board[0])
        uf = Solution.UnionFind(m*n+1)
        dummy = m*n
        for i in range(m):
            for j in range(n):
                if board[i][j] == 'O':
                    if i == 0 or i == m-1 or j == 0 or j == n-1:
                        uf.union(i*n+j,dummy)  
                    else:
                        if i>0 and board[i-1][j] == 'O':
                            uf.union(i*n+j,(i-1)*n+j)
                        if i < m - 1 and board[i + 1][j] == 'O':
                            uf.union(i*n+j, (i+1)*n+j)
                        if j > 0 and board[i][j - 1] == 'O':
                            uf.union(i*n+j, i*n+j-1)
                        if j < n - 1 and board[i][j + 1] == 'O':
                            uf.union(i*n+j, i*n+j+1)           
        for i in range(m):
            for j in range(n):
                if uf.isConnected(i*n+ j, dummy):
                    #和dummyNode 在一个连通区域的,那么就是O；
                    board[i][j] = 'O'
                else:
                    board[i][j] = 'X'
                

990、判定合法等式

具体实现：

如果是等号就连接两个节点，

如果是不等号就判断这两个节点是否是同一个根节点

是的话就判断为False

代码：

class Solution:

    class UnionFind:
        def __init__(self):
            self.parent = list(range(26))
        
        def find(self, index):
            if index == self.parent[index]:
                return index
            self.parent[index] = self.find(self.parent[index])
            return self.parent[index]
        
        def union(self, index1, index2):
            self.parent[self.find(index1)] = self.find(index2)


    def equationsPossible(self, equations: List[str]) -> bool:
        uf = Solution.UnionFind()
        for st in equations:
            if st[1] == "=":#如果是等于号就连接两个节点
                index1 = ord(st[0]) - ord("a")
                index2 = ord(st[3]) - ord("a")
                uf.union(index1, index2)
        for st in equations:
            if st[1] == "!":#如果是不等于号
                index1 = ord(st[0]) - ord("a")
                index2 = ord(st[3]) - ord("a")
                if uf.find(index1) == uf.find(index2):#判断这两个节点是否是一个根节点
                    return False #是的话就放回False
        return True