面试题26：复杂链表的复制

分解复杂问题简单化

将问题分为3步，参考P147

结构：

struct ComplexListNode
{
    int                 m_nValue;
    ComplexListNode*    m_pNext;
    ComplexListNode*    m_pSibling;
};

1：复制节点，让N'在N后行成一条新链表

 1 void CloneNodes(ComplexListNode* pHead)
 2 {
 3     ComplexListNode* pNode = pHead;
 4     while(pNode != NULL)
 5     {
 6         ComplexListNode* pCloned = new ComplexListNode();
 7         pCloned->m_nValue = pNode->m_nValue;
 8         pCloned->m_pNext = pNode->m_pNext;
 9         pCloned->m_pSibling = NULL;
10  
11         pNode->m_pNext = pCloned;
12  
13         pNode = pCloned->m_pNext;
14     }
15 }

2：遍历新链表，设置N'的m_pSibling的值

 1 void ConnectSiblingNodes(ComplexListNode* pHead)
 2 {
 3     ComplexListNode* pNode = pHead;
 4     while(pNode != NULL)
 5     {
 6         ComplexListNode* pCloned = pNode->m_pNext;
 7         if(pNode->m_pSibling != NULL)
 8         {
 9             pCloned->m_pSibling = pNode->m_pSibling->m_pNext;
10         }
11  
12         pNode = pCloned->m_pNext;
13     }
14 }

3：将新链表中的N和N'分离

 1 ComplexListNode* ReconnectNodes(ComplexListNode* pHead)
 2 {
 3     ComplexListNode* pNode = pHead;
 4     ComplexListNode* pClonedHead = NULL;
 5     ComplexListNode* pClonedNode = NULL;
 6  
 7     if(pNode != NULL)
 8     {
 9         pClonedHead = pClonedNode = pNode->m_pNext;
10         pNode->m_pNext = pClonedNode->m_pNext;
11         pNode = pNode->m_pNext;
12     }
13  
14     while(pNode != NULL)
15     {
16         pClonedNode->m_pNext = pNode->m_pNext;
17         pClonedNode = pClonedNode->m_pNext;
18  
19         pNode->m_pNext = pClonedNode->m_pNext;
20         pNode = pNode->m_pNext;
21     }
22  
23     return pClonedHead;
24 }

另外一种方法，用map

http://www.nowcoder.com/practice/f836b2c43afc4b35ad6adc41ec941dba?tpId=13&tqId=11178&rp=2&ru=%2Fta%2Fcoding-interviews&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking

 1 /*
 2 struct RandomListNode {
 3     int label;
 4     struct RandomListNode *next, *random;
 5     RandomListNode(int x) :
 6             label(x), next(NULL), random(NULL) {
 7     }
 8 };
 9 */
10 /*
11 首先遍历一遍原链表，创建新链表（赋值label和next），用map关联对应结点；再遍历一遍，更新新链表的random指针。（注意map中应有NULL ----> NULL的映射）
12 */
13 class Solution {
14 public:
15     RandomListNode* Clone(RandomListNode* pHead)
16     {
17         if(pHead==NULL) return NULL;
18  
19         map<RandomListNode*,RandomListNode*> m;
20         RandomListNode* pHead1 = pHead;
21         RandomListNode* pHead2 = new RandomListNode(pHead1->label);
22         RandomListNode* newHead = pHead2;
23         m[pHead1] = pHead2;
24         while(pHead1){
25             if(pHead1->next) pHead2->next = new RandomListNode(pHead1->next->label);
26             else pHead2->next = NULL;
27             pHead1 = pHead1->next;
28             pHead2 = pHead2->next;
29             m[pHead1] = pHead2;
30         }
31  
32         pHead1 = pHead;
33         pHead2 = newHead;
34         while(pHead1){
35             pHead2->random = m[pHead1->random];
36             pHead1 = pHead1->next;
37             pHead2 = pHead2->next;
38         }
39         return newHead;
40     }
41 };

posted on 2016-07-08 09:12 已停更阅读(196) 评论(0) 收藏举报