刷题笔记 - 0422

Posted 2020-11-01 li-ling

1、clone graph   用map结构存储原值和拷贝值，一一对应。

map红黑树实现，O(logn) 内部有序；每个节点要保存父节点、子节点及红黑属性，占用空间大。

unordered_map哈希表实现，查找O(1)，内部无序；哈希表建立耗时。   常用unordered_map

class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        // 拷贝所有点  DFS写法
        if (!node)  return nullptr;
        unordered_map<int, UndirectedGraphNode *> table;
        return helper(node, table);
    }
    
    UndirectedGraphNode *helper(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode*>& table) {
        if (!node)  return nullptr;
        // 如果表中已有数据
        if (table.find(node->label) != table.end()) {
            return table[node->label];
        }
        UndirectedGraphNode *newNode = new UndirectedGraphNode(node->label);
        table[node->label] = newNode;
        for (int i = 0; i < node->neighbors.size(); i++) {
            UndirectedGraphNode *tmp = helper(node->neighbors[i], table);
            newNode->neighbors.push_back(tmp);
        }
        return newNode;
    }
};

 

 1 /**
 2  * Definition for undirected graph.
 3  * struct UndirectedGraphNode {
 4  *     int label;
 5  *     vector<UndirectedGraphNode *> neighbors;
 6  *     UndirectedGraphNode(int x) : label(x) {};
 7  * };
 8  */
 9 class Solution {
10 public:
11     UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
12         // 拷贝所有点  用BFS或DFS遍历都可以  是否已拷贝用map判断
13         if (!node)  return nullptr;
14         unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> map;
15         UndirectedGraphNode *p1 = node;
16         UndirectedGraphNode *p2 = new UndirectedGraphNode(node->label);
17         queue<UndirectedGraphNode *> q;
18         q.push(node);
19         map[p1] = p2;
20         while (!q.empty()) {
21             p1 = q.front();   q.pop();
22             p2 = map[p1];
23             for (int i = 0; i < p1->neighbors.size(); i++) {
　　　　　　　　// 注意对邻接点的判断 若map中已有，则直接设置p2的邻接点；否则插入map，入队列处理
24                 UndirectedGraphNode *tmp = p1->neighbors[i];
25                 if (map.count(tmp) != 0) {
26                     p2->neighbors.push_back(map[tmp]);
27                 } else {
28                     UndirectedGraphNode *copy = new UndirectedGraphNode(tmp->label);
29                     map[tmp] = copy;
30                     p2->neighbors.push_back(map[tmp]);
31                     q.push(tmp);
32                 }
33             }
34         }
35         return map[node];
36     }
37 };

2、二叉树序列化    https://leetcode.com/problems/serialize-and-deserialize-bst/discuss/93167/Concise-C++-19ms-solution-beating-99.4

队列的数组实现与链表实现，用vector实现queue