[Leetcode学习-java]Clone Graph(地图深克隆)

问题：

难度：medium

说明：

对一个相互依赖的对象，进行一次深克隆，返回一个全新的对象。

问题链接：https://leetcode.com/problems/clone-graph/

输入案例：

Input: adjList = [[2,4],[1,3],[2,4],[1,3]] Output: [[2,4],[1,3],[2,4],[1,3]] Explanation: There are 4 nodes in the graph. 1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4). 2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3). 3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4). 4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

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我的代码：

其实这个和spring的解决依赖循环一个意思，也跟 tow sum 一样的解决方法，使用一个 map 进行缓存，然后进行递归实例化未实例化的对象，再进行 neighbors 添加。

public class Solution { public Node cloneGraph(Node node) { if (node == null) { return node; } // 搞个缓存 HashMap<Integer, Node> cacheMap = new HashMap<Integer, Node>(); return clone(node, cacheMap); } public Node clone(Node node, HashMap<Integer, Node> cacheMap) { Node newNode = new Node(); newNode.val = node.val; newNode.neighbors = new ArrayList<>(); cacheMap.put(node.val, newNode); for(Node ne : node.neighbors) { if(cacheMap.containsKey(ne.val)) { newNode.neighbors.add(cacheMap.get(ne.val)); } else { Node one = clone(ne, cacheMap); // 递归逐个实例化 one.val = ne.val; newNode.neighbors.add(one); } } return newNode; } }