数据结构与算法-并查集

Posted by 小拳头 on Monday, February 15, 2021

框架

照搬的参考3.

class UF {
    private int count; //记录连通分量个数
    private int[] parent; //存储若干棵树
    private int[] size; //记录树的大小

    public UF(int n) {
        this.count = n;
        parent = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            size[i] = 1;
        }
    }

    //将p和q连通
    public void union(int p, int q) {
        int rootP = find(p);
        int rootQ = find(q);
        if (rootP == rootQ)
            return;

        //小树接到大树下面
        if (size[rootP] > size[rootQ]) {
            parent[rootQ] = rootP;
            size[rootP] += size[rootQ];
        } else {
            parent[rootP] = rootQ;
            size[rootQ] += size[rootP];
        }
        count--;
    }

    //判断p和q是否互相连通
    public boolean connected(int p, int q) {
        int rootP = find(p);
        int rootQ = find(q);
        // 处于同一棵树上的节点相互连通
        return rootP == rootQ;
    }

    //返回节点x的根节点
    private int find(int x) {
        //进行路径压缩
        while (parent[x] != x) {
            parent[x] = parent[parent[x]];
            x = parent[x];
        }
        return x;
    }

    public int count() {
        return count;
    }
}

547. 省份数量

用没有优化的并查集解决.

class Solution {
    int[] parent;
    int count;
    public int findCircleNum(int[][] isConnected) {
        count = isConnected.length;
        parent = new int[isConnected.length];
        for (int i = 0; i < isConnected.length; i++) parent[i] = i;
        for (int i = 0; i < isConnected.length; i++) {
            for (int j = 0; j < isConnected.length; j++) {
                if (i != j && isConnected[i][j] == 1)
                    union(i, j);
            }
        }
        return count;
    }

    public void union(int p, int q) {
        int rootP = find(p);
        int rootQ = find(q);
        if (rootP == rootQ) return;
        parent[rootP] = rootQ;
        count--;
    }

    public int find(int x) {
        while (parent[x] != x)
            x = parent[x];
        return x;
    }
}

参考

  1. labuladong算法
  2. labuladong-并查集
  3. labuladong-并查集
  4. leetcode

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