定义

顾名思义（看英文），就是并集，查找，即找到父节点，然后取并集；

题目：LeetCode 547. 省份数量
解决方法：参照B站Up主，塔罗兔

代码：

• 基础版本

class Solution {
public:
    vector<int> fathers;
    int find(int i) {
        while (fathers[i] != i) {
            i = fathers[i];
        }
        return i;
    }
    void uni(int i, int j) {
        int f1 = find(i);
        int f2 = find(j);
        fathers[f2] = f1;
    }
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        // 初始化父节点
        for (int i = 0; i < n; i++) {
            fathers.push_back(i);
        }
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (isConnected[i][j]) {
                    uni(i, j);
                }
            }
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (fathers[i] == i) count++;
        }
        return count;
    }
};

• 优化：

class Solution {
public:
    vector<int> fathers;
    vector<int> ranks;
    int find(int i) {
        while (fathers[i] != i) {
            i = fathers[i];
        }
        return i;
    }
    void uni(int i, int j) {
        int f1 = find(i);
        int f2 = find(j);
        // fathers[f2] = f1;
        if (ranks[f1] == ranks[f2]) {
            fathers[f2] = f1;
            ranks[f1]++;
        } else if (ranks[f1] > ranks[f2]) {
            fathers[f2] = f1;
        } else {
            fathers[f1] = f2;
        }

    }
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        // 初始化父节点
        for (int i = 0; i < n; i++) {
            fathers.push_back(i);
            ranks.push_back(1);
        }
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (isConnected[i][j]) {
                    uni(i, j);
                }
            }
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (fathers[i] == i) count++;
        }
        return count;
    }
};

PS: 代码运行的时间反而增加了，在LeetCode[CN] 版本中.

标签：f1,f2,Union,fathers,查集,++,int,ranks,Find
来源： https://www.cnblogs.com/ache/p/16519379.html