刷题138. Copy List with Random Pointer

一、题目说明

题目138. Copy List with Random Pointer，一个链表中的节点包括一个random指针，可以指向链表中的任何节点或者空，生成该链表的一个拷贝。难度是Medium！

二、我的解答

这个题目是赋值一个链表，难度在于random链表的指向。这里用计数方式实现：

class Solution{
	public:
		Node* copyRandomList(Node* head){
			if(head == NULL) return NULL;
			
			Node* p = head->next,*newHead,*newP,*cur;
			newHead = new Node(head->val);
			newP = newHead;
			
			
			while(p!=NULL){
				newP->next = new Node(p->val);
				newP = newP->next;
				p = p->next;
			}
			
			p = head;
			newP = newHead;
			int n = 0;
			while(p!=NULL){
				cur = p->random;
				if(cur!=NULL){
					//从head开始遍历，数第几个 
					Node* p1 = head;
					while(p1 !=cur){
						p1 = p1->next;
						n++;
					}
					
					Node*p2 = newHead;
					while(n>0){
						p2 = p2->next;
						n--;
					}
					newP->random = p2;
				}
				
				p = p->next;
				newP = newP->next;
			}
			
			return newHead;
		}
};

性能如下：

Runtime: 12 ms, faster than 61.45% of C++ online submissions for Copy List with Random Pointer.
Memory Usage: 13.4 MB, less than 100.00% of C++ online submissions for Copy List with Random Pointer.

三、优化措施

由于random指针用的是“从头到尾”计数的方式实现，性能一般。可以用Hash方式加速，而且更容易理解：

class Solution{
	public:
		Node* copyRandomList(Node* head){
			if(head == NULL) return NULL;
			Node* cur = head;
			unordered_map<Node*,Node*> ump;
			//拷贝节点，复制val 
			while(cur!=NULL){
				Node* copy = new Node(cur->val);
				ump[cur] = copy;
				cur = cur->next;
			} 
			//复制next和random
			cur = head;
			while(cur!=NULL){
				ump[cur]->next = ump[cur->next];
				ump[cur]->random = ump[cur->random];
				cur = cur->next;
			}
			
			return ump[head];
		}
};

性能如下：

Runtime: 8 ms, faster than 89.21% of C++ online submissions for Copy List with Random Pointer.
Memory Usage: 13.5 MB, less than 100.00% of C++ online submissions for Copy List with Random Pointer.