【Leetcode】并查集

990. 等式方程的可满足性

class Solution {
    public boolean equationsPossible(String[] equations) {
        int[] parent = new int[26];//存放节点的父节点
        for(int i = 0; i < 26; i++){
            parent[i] = i;//初始化自己是自己的父节点
        }
        for(String equation: equations){
            int index1 = equation.charAt(0)-'a';
            int index2 = equation.charAt(3)-'a';
            if(equation.charAt(1) == '='){//相等则合并在一个连通集合中
                union(parent, index1, index2);
            }
        }
        for(String equation: equations){
            if(equation.charAt(1) == '!'){//不等但是在一个连通集合中 说明矛盾
                int parent1 = find(parent, equation.charAt(0)-'a');
                int parent2 = find(parent,equation.charAt(3)-'a');
                if(parent1 == parent2){//顶层节点相同
                    return false;
                }
            }
        }
        return true;
    }
    private void union(int[] parent, int index1, int index2){
        parent[find(parent,index1)] = find(parent,index2);//index2的父等于index1的父
    }
    private int find(int[] parent, int index){//查找顶层节点
        while(parent[index] != index){
            index = parent[index];//一直向上找，直到找到顶层父
        }
        //如果节点的父节点是自己则是顶层节点 直接返回
        return index;
    }
}