574.Number of Provinces

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 directfriend 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 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:

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

link：https://leetcode-cn.com/problems/number-of-provinces

解题思路

1. 深度优先搜索

   遍历节点，如果访问过则跳到下一个，如果没访问过，结果增1，

   遍历到节点时，对每个节点再遍历其关系，如果是有关系的则递归遍历。如果访问过，则标记一下。

2. 广度优先搜索

3. 并查集

代码

/** 深度优先搜索
 * @param {number[][]} isConnected
 * @return {number}
 */
var mark = function(i, isConnected, markedList) {
    markedList[i] = 1;
    var city = isConnected[i];
    var n = city.length;
    for (var j = 0; j < n; j++) {
        if (city[j]) {
            if (i !== j && markedList[j] !== 1) {
                mark(j, isConnected, markedList);
            }
        }
    }
}
var findCircleNum = function(isConnected) {
    var n = isConnected.length;
    var markedList = new Array(n).fill(0); // 该城市是否被标记
    var result = 0;
    for (var i = 0; i < n; i++) {
        //如果已经标记 则推出
        if (markedList[i] === 1) {
            continue;
        }
        result+=1;
        mark(i, isConnected, markedList);
    }
    return result;
};