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

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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]

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想象一下你是個城市基建規劃者，地圖上有 N 座城市，它們按以 1 到 N 的次序編號。

給你一些可鏈接的選項 conections，其中每一個選項 conections[i] = [city1, city2, cost] 表示將城市 city1 和城市 city2 鏈接所要的成本。（鏈接是雙向的，也就是說城市 city1 和城市 city2 相連也一樣意味着城市 city2 和城市 city1 相連）。

返回使得每對城市間都存在將它們鏈接在一塊兒的連通路徑（可能長度爲 1 的）最小成本。該最小成本應該是所用所有鏈接代價的綜合。若是根據已知條件沒法完成該項任務，則請你返回 -1。

示例 1：

輸入：N = 3, conections = [[1,2,5],[1,3,6],[2,3,1]]
輸出：6
解釋：
選出任意 2 條邊均可以鏈接全部城市，咱們從中選取成本最小的 2 條。

示例 2：

輸入：N = 4, conections = [[1,2,3],[3,4,4]]
輸出：-1
解釋： 
即便連通全部的邊，也沒法鏈接全部城市。

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]