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An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1
) is given as graph
.node
graph.length = N
, and j != i
is in the list graph[i]
exactly once, if and only if nodes i
and j
are connected.git
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.github
Example 1:微信
Input: [[1,2,3],[0],[0],[0]] Output: 4 Explanation: One possible path is [1,0,2,0,3]
Example 2:app
Input: [[1],[0,2,4],[1,3,4],[2],[1,2]] Output: 4 Explanation: One possible path is [0,1,4,2,3]
Note:spa
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。 code
graph.length = N
,且只有节点 i
和 j
连通时,j != i
在列表 graph[i]
中刚好出现一次。htm
返回可以访问全部节点的最短路径的长度。你能够在任一节点开始和中止,也能够屡次重访节点,而且能够重用边。blog
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 }