【题35 复杂链表复制】

阳光穿透心脏的1/2处 2022-03-09 00:48 173阅读 0赞

【题目】输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的head。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）  
【思路】  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zOTc5NTA0OQ_size_16_color_FFFFFF_t_70]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zOTc5NTA0OQ_size_16_color_FFFFFF_t_70 1]  
【实现】

/*
    public class RandomListNode {
        int label;
        RandomListNode next = null;
        RandomListNode random = null;
    
        RandomListNode(int label) {
            this.label = label;
        }
    }
    */
    public class Solution {
        public RandomListNode Clone(RandomListNode pHead){
            if(pHead == null){
                return null;
            }
            //第一步
            RandomListNode currentNode = pHead;
            while(currentNode != null){
                RandomListNode cloneNode = new RandomListNode(currentNode.label);
                RandomListNode nextNode = currentNode.next;
                currentNode.next = cloneNode;
                cloneNode.next = nextNode;
                currentNode = nextNode;
            }
            //第二步
            currentNode = pHead;
            while(currentNode != null){
                currentNode.next.random = currentNode.random == null?null:currentNode.random.next;
                currentNode = currentNode.next.next;
            }
            //第三步
            currentNode = pHead;
            RandomListNode pCloneHead = pHead.next;
            while(currentNode != null){
                RandomListNode cloneNode = currentNode.next;
                currentNode.next = cloneNode.next;
                cloneNode.next = cloneNode.next == null?null:cloneNode.next.next;
                currentNode = currentNode.next;
            }
            return pCloneHead;
        }
    }

【代码解析】  
第一步：复制每个结点，如复制结点A得到A‘，将结点A’插到结点A后面；

RandomListNode currentNode = pHead;
            while(currentNode != null){
                RandomListNode cloneNode = new RandomListNode(currentNode.label);
                RandomListNode nextNode = currentNode.next;//保存一个节点nextNode
                currentNode.next = cloneNode;//将cloneNode连在currentNode后面
                cloneNode.next = nextNode;//将保存的nextNode节点连在复制的节点的后面
                currentNode = nextNode;
            }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zOTc5NTA0OQ_size_16_color_FFFFFF_t_70 2]  
第二步：重新遍历链表，复制老结点的随机指针给新结点，如A1.random = A.random.next;

currentNode = pHead;
            while(currentNode != null){
                currentNode.next.random = currentNode.random == null?null:currentNode.random.next;
                currentNode = currentNode.next.next;
            }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zOTc5NTA0OQ_size_16_color_FFFFFF_t_70 3]  
第三步：拆分链表，将链表拆分为原链表和复制后的链表

currentNode = pHead;
            RandomListNode pCloneHead = pHead.next;
            while(currentNode != null){
                RandomListNode cloneNode = currentNode.next;
                currentNode.next = cloneNode.next;
                cloneNode.next = cloneNode.next == null?null:cloneNode.next.next;
                currentNode = currentNode.next;
            }

参考：  
1.《剑指offer》

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