[Swift]LeetCode1135. 最低成本联通全部城市 | Connecting Cities With Minimum Cost

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-zxgnceqp-ku.html 
➤若是连接不是山青咏芝的博客园地址，则多是爬取做者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持做者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

There are N cities numbered from 1 to N.

You are given connections, where each connections[i] = [city1, city2, cost] represents the cost to connect city1 and city2 together.  (A connection is bidirectional: connecting city1 and city2is the same as connecting city2 and city1.)

Return the minimum cost so that for every pair of cities, there exists a path of connections (possibly of length 1) that connects those two cities together.  The cost is the sum of the connection costs used. If the task is impossible, return -1.

 

Example 1:

Input: N = 3, connections = [[1,2,5],[1,3,6],[2,3,1]]
Output: 6
Explanation: 
Choosing any 2 edges will connect all cities so we choose the minimum 2.

Example 2:

Input: N = 4, connections = [[1,2,3],[3,4,4]]
Output: -1
Explanation: 
There is no way to connect all cities even if all edges are used.

Note:

1. 1 <= N <= 10000
2. 1 <= connections.length <= 10000
3. 1 <= connections[i][0], connections[i][1] <= N
4. 0 <= connections[i][2] <= 10^5
5. connections[i][0] != connections[i][1]

---

想象一下你是个城市基建规划者，地图上有 N 座城市，它们按以 1 到 N 的次序编号。

给你一些可链接的选项 conections，其中每一个选项 conections[i] = [city1, city2, cost] 表示将城市 city1 和城市 city2 链接所要的成本。（链接是双向的，也就是说城市 city1 和城市 city2 相连也一样意味着城市 city2 和城市 city1 相连）。

返回使得每对城市间都存在将它们链接在一块儿的连通路径（可能长度为 1 的）最小成本。该最小成本应该是所用所有链接代价的综合。若是根据已知条件没法完成该项任务，则请你返回 -1。

示例 1：

输入：N = 3, conections = [[1,2,5],[1,3,6],[2,3,1]]
输出：6
解释：
选出任意 2 条边均可以链接全部城市，咱们从中选取成本最小的 2 条。

示例 2：

输入：N = 4, conections = [[1,2,3],[3,4,4]]
输出：-1
解释： 
即便连通全部的边，也没法链接全部城市。

提示：

1. 1 <= N <= 10000
2. 1 <= conections.length <= 10000
3. 1 <= conections[i][0], conections[i][1] <= N
4. 0 <= conections[i][2] <= 10^5
5. conections[i][0] != conections[i][1]