【leetcode】990. Satisfiability of Equality Equations

题目如下：

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 bare 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 '='

解题思路：我的方法是用并查集，先把所有等式中的字母合并组成并查集，接下来再判断不等式中的字母是否属于同一祖先。

代码如下：

class Solution(object):
    def equationsPossible(self, equations):
        """
        :type equations: List[str]
        :rtype: bool
        """
        def find(v):
            if parent[v] == v:
                return v
            return find(parent[v])

        def union(v1,v2):
            p1 = find(v1)
            p2 = find(v2)
            if p1 < p2:
                parent[p2] = p1
            else:
                parent[p1] = p2
        parent = [i for i in range(26)]

        uneuqal = []
        while len(equations) > 0:
            item = equations.pop(0)
            if item[1] == '!':
                uneuqal.append(item)
            else:
                union(ord(item[0]) - ord('a'),ord(item[3]) - ord('a'))

        for item in uneuqal:
            if find(ord(item[0]) - ord('a')) == find(ord(item[3]) - ord('a')):
                return False
        return True