990. Satisfiability of Equality Equations

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Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily different) that represent one-letter variable names.

Return true if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.

 

Example 1:

Input: ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.

Example 2:

Input: ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.

Example 3:

Input: ["a==b","b==c","a==c"]
Output: true

Example 4:

Input: ["a==b","b!=c","c==a"]
Output: false

Example 5:

Input: ["c==c","b==d","x!=z"]
Output: true

 

Note:

1. 1 <= equations.length <= 500
2. equations[i].length == 4
3. equations[i][0] and equations[i][3] are lowercase letters
4. equations[i][1] is either '=' or '!'
5. equations[i][2] is '='

思路：先根据等号建立union-find，然后对不等号进行find，如果find结果相同说明冲突

class Solution(object):
    def equationsPossible(self, equations):
        """
        :type equations: List[str]
        :rtype: bool
        """
        fa=[i for i in range(26)]
        base=ord('a')
        def find(x):
            while x!=fa[x]: x=fa[x]
            return x
        def union(i,j):
            ii,jj=find(i),find(j)
            fa[ii]=jj
        for eq in equations:
            if eq[1]=='=':
                union(ord(eq[0])-base, ord(eq[3])-base)
        
        for eq in equations:
            if eq[1]=='!':
                if find(ord(eq[0])-base)==find(ord(eq[3])-base):
                    return False
        return True