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➤微信公众号:山青咏芝(shanqingyongzhi)
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An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1
) is given as graph
.
graph.length = N
, and j != i
is in the list graph[i]
exactly once, if and only if nodes i
and j
are connected.
Return the length of the shortest path that visits every node. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.
Example 1:
Input: [[1,2,3],[0],[0],[0]] Output: 4 Explanation: One possible path is [1,0,2,0,3]
Example 2:
Input: [[1],[0,2,4],[1,3,4],[2],[1,2]] Output: 4 Explanation: One possible path is [0,1,4,2,3]
Note:
1 <= graph.length <= 12
0 <= graph[i].length < graph.length
给出 graph
为有 N 个节点(编号为 0, 1, 2, ..., N-1
)的无向连通图。
graph.length = N
,且只有节点 i
和 j
连通时,j != i
在列表 graph[i]
中恰好出现一次。
返回能够访问所有节点的最短路径的长度。你可以在任一节点开始和停止,也可以多次重访节点,并且可以重用边。
示例 1:
输入:[[1,2,3],[0],[0],[0]] 输出:4 解释:一个可能的路径为 [1,0,2,0,3]
示例 2:
输入:[[1],[0,2,4],[1,3,4],[2],[1,2]] 输出:4 解释:一个可能的路径为 [0,1,4,2,3]
提示:
1 <= graph.length <= 12
0 <= graph[i].length < graph.length
1 class Solution { 2 func shortestPathLength(_ graph: [[Int]]) -> Int { 3 var n:Int = graph.count 4 var fullMask:Int = (1 << n) - 1 5 var visited:Set<String> = Set<String>() 6 var que:[Node] = [Node]() 7 for i in 0..<n 8 { 9 var node:Node = Node(i, 1<<i) 10 que.append(node) 11 visited.insert(node.toString()) 12 } 13 var level:Int = 0 14 while(!que.isEmpty) 15 { 16 var size:Int = que.count 17 for i in 0..<size 18 { 19 var node = que.removeFirst() 20 if node.mask == fullMask {return level} 21 for next in graph[node.id] 22 { 23 var nextNode:Node = Node(next, node.mask | (1 << next)) 24 if visited.contains(nextNode.toString()) {continue} 25 que.append(nextNode) 26 visited.insert(nextNode.toString()) 27 } 28 } 29 level += 1 30 } 31 return level 32 } 33 } 34 35 class Node { 36 var id:Int = 0 37 var mask:Int = 0 38 39 init(_ id:Int,_ mask:Int) 40 { 41 self.id = id 42 self.mask = mask 43 } 44 45 func toString() -> String 46 { 47 return String(id) + " " + String(mask) 48 } 49 }