Friend Circles

最新推荐文章于 2020-06-15 09:07:48 发布

flyatcmu

最新推荐文章于 2020-06-15 09:07:48 发布

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分类专栏： Union-Find

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本文介绍了一种使用并查集(Union Find)算法来计算学生之间的朋友圈数量的方法。通过输入一个N*N的矩阵来表示学生之间的直接友谊关系，该算法能够找出所有间接朋友形成的朋友圈，并给出朋友圈的总数。

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jthstudents are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:

Input: 
[[1,1,0],
 [1,1,0],
 [0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
The 2nd student himself is in a friend circle. So return 2.

Example 2:

Input: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.

Note:

N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

思路：刚开始以为这题跟Number of Island 一样，后来测试之后发现不是一个东西。比如

[

1 0 0 1

0 1 1 0

1 0 1 1

]

如果number of island 返回4，然而这题返回1.也就是说，number island是探索孤立的岛屿，而这题是朋友圈，比如1，3认识，3跟2认识，2跟1认识，他们所有人的都在一个朋友圈里面，相互传染了，这题就是为了union find设计的。

思路：标准UnionFind模板，解决；

class Solution {
    
    private class UnionFind {
        private int[] father;
        private int count;
        public UnionFind(int n) {
            father = new int[n+1];
            for(int i = 0; i <= n; i++) {
                father[i] = i;
            }
            this.count = n;
        }
        
        public int find(int x) {
            int j = x;
            while(j != father[j]) {
                j = father[j];
            }
            
            // path compression;
            while(x != j) {
                int fx = father[x];
                father[x] = j;
                x = fx;
            }
            
            return j;
        }
        
        public void union(int a, int b) {
            int father_a = find(a);
            int father_b = find(b);
            if(father_a != father_b) {
                father[father_a] = father_b;
                this.count--;
            }
        }
        
        public int getCount() {
            return this.count;
        }
    }
    
    public int findCircleNum(int[][] M) {
        if(M == null || M.length == 0 || M[0].length == 0){
            return 0;
        }
        int n = M.length;
        UnionFind uf = new UnionFind(n);
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(M[i][j] == 1) {
                    uf.union(i,j);
                }
            }
        }
        return uf.getCount();
    }
}