LeetCode Friend Circles 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 jth students are directfriends 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.
给定一个矩阵，M[i][j]=1表示i和j有直接关系，如果i和j有直接关系，j和k有直接关系，则i和k有间接关系。问这个矩阵共有多少个关系网。 简单题，有两种解法。第一种解法是DFS或者BFS，每次把能搜索到的点标记为一个新的关系网，直到所有点都属于一个关系网。但是无论DFS还是BFS，复杂度都比较高。 这种题应该条件反射想到并查集，只要M[i][j]=1，则把i和j union起来，最后看一下有多少个代表就好了。这种解法非常简单，代码如下： [cpp] class Solution { private: int n; vector<int> parent; void init() { parent.resize(n); for (int i = 0; i < n; ++i) { parent[i] = i; } } int find_set(int x) { if (parent[x] != x) parent[x] = find_set(parent[x]); return parent[x]; } void union_set(int u, int v) { parent[find_set(u)] = find_set(v); } public: int findCircleNum(vector<vector<int>>& M) { n = M.size(); init(); for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { if (M[i][j] == 1) { union_set(i, j); } } } set<int> circles; for (int i = 0; i < n; ++i)circles.insert(find_set(i)); // 注意最后还要find一下找到代表 return circles.size(); } }; [/cpp] 本代码提交AC，用时19MS。]]>