|
| 1 | +/** |
| 2 | + * res.js |
| 3 | + * @authors Joe Jiang (hijiangtao@gmail.com) |
| 4 | + * @date 2017-04-17 14:02:24 |
| 5 | + * |
| 6 | + * For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels. |
| 7 | + * |
| 8 | + * Format |
| 9 | + * The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels). |
| 10 | + * |
| 11 | + * You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges. |
| 12 | + * |
| 13 | + * Note: |
| 14 | + * |
| 15 | + * (1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.” |
| 16 | + * |
| 17 | + * (2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf. |
| 18 | + * |
| 19 | + * @param {number} n |
| 20 | + * @param {number[][]} edges |
| 21 | + * @return {number[]} |
| 22 | + */ |
| 23 | +let findMinHeightTrees = function(n, edges) { |
| 24 | + let elen = edges.length, // 边数长度 |
| 25 | + nlist = [], // 节点列表 |
| 26 | + deglist = [], //度数列表 |
| 27 | + adj = new Array(n); //存储连边信息 |
| 28 | + |
| 29 | + for (let i = 0; i < n; i++) { |
| 30 | + nlist.push(i); |
| 31 | + deglist.push(0); |
| 32 | + adj[i] = new Set(); |
| 33 | + } |
| 34 | + for (let i = 0; i < elen; i++) { |
| 35 | + let source = edges[i][0], |
| 36 | + target = edges[i][1]; |
| 37 | + |
| 38 | + adj[source].add(target); |
| 39 | + adj[target].add(source); |
| 40 | + deglist[source]++; |
| 41 | + deglist[target]++; |
| 42 | + } |
| 43 | + |
| 44 | + // 结果中只能是一个元素或者两个元素, 或者全部元素 (如果有多个树结构) |
| 45 | + while (nlist.length > 2) { |
| 46 | + let lenNow = nlist.length, |
| 47 | + dellist = []; |
| 48 | + |
| 49 | + for (let i = 0; i < lenNow; i++) { |
| 50 | + let node = nlist[i]; |
| 51 | + if (!deglist[node]) { |
| 52 | + //当前节点边数为0 |
| 53 | + nlist.splice(i--, 1); |
| 54 | + lenNow--; |
| 55 | + } else if (deglist[node] === 1) { |
| 56 | + //删除边并减少两端节点的degree |
| 57 | + let anothernode = -1; |
| 58 | + for (let j = 0; j < lenNow; j++) { |
| 59 | + if (i === j) continue; |
| 60 | + if (adj[node].has(nlist[j])) { |
| 61 | + anothernode = nlist[j]; |
| 62 | + break; |
| 63 | + } |
| 64 | + } |
| 65 | + adj[node].delete(anothernode); |
| 66 | + adj[anothernode].delete(node); |
| 67 | + dellist.push(anothernode); |
| 68 | + deglist[node] = 0; |
| 69 | + nlist.splice(i--, 1); |
| 70 | + lenNow--; |
| 71 | + } |
| 72 | + } |
| 73 | + |
| 74 | + for (let i = dellist.length - 1; i >= 0; i--) { |
| 75 | + deglist[dellist[i]]--; |
| 76 | + } |
| 77 | + } |
| 78 | + |
| 79 | + return nlist; |
| 80 | +}; |
0 commit comments