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[LeetCode] Intersection of Two Linked Lists 两链表是否相交

Write a program to find the node at which the intersection of two singly linked lists begins.


For example, the following two linked lists:

A:          a1 → a2                   ↘                     c1 → c2 → c3                   ↗            B:     b1 → b2 → b3

begin to intersect at node c1.


Notes:

  • If the two linked lists have no intersection at all, return null.
  • The linked lists must retain their original structure after the function returns.
  • You may assume there are no cycles anywhere in the entire linked structure.
  • Your code should preferably run in O(n) time and use only O(1) memory.

 

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

 

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 Linked List
 
     两个单项链表,判断是否存在交集,如上图很清晰,最直观的方法是
for  list1 begin to last
  for list2 begin to last
    if list2==list1 success
  end
end  
    时间是O(nm),空间挺少的O(1)。如何提高呢?
  1. 遍历list1 ,将其节点存在hash_table
  2. 遍历list2,如果已经在hash_table中,那么存在

    利用hash_table 可以提升时间到O(n+m),可是空间变O(n)了

 1 class Solution { 2 public: 3     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { 4         unordered_map<ListNode*,int> m; 5         while(headA!=NULL){ 6             m[headA] = 1; 7             headA=headA->next; 8         } 9         while(headB!=NULL){10             if(m[headB]==1) return headB;11             headB=headB->next;12         }13         return NULL;14     }15 };
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  那题目的最好解法,这技巧问题阿,遍历list1 后接着遍历list2,同时,遍历list2然后遍历list1,这样两个遍历的长度是一样的O(n+m),怎么判断相等呢?
 
list1:    O O O O O ⑴ ⑵ ⑶
list2:    □ □ □ □ ⑴ ⑵ ⑶
  假如list 如上,⑴ ⑵ ⑶ 为相同的节点,那么遍历list1 这样便是这样:
O O O O O ⑴ ⑵ ⑶ □ □ □ □ ⑴ ⑵ ⑶
  遍历list2 便是这样。
□  □ □ □ ⑴ ⑵ ⑶ O O O O O ⑴ ⑵ ⑶
 
合在一起看看:
O  O  O  O  O  ⑴  ⑵  ⑶  □   □  □  □   ⑴  ⑵  ⑶
□   □  □   □  ⑴  ⑵  ⑶  O  O  O  O  O  ⑴  ⑵  ⑶
 
    好了,现在规律出来了。这个逻辑出来明显,主要麻烦是在遍历一个结束后接上第二个,直接改链表不好,所以,使用flag 控制。
算法逻辑:
  1. 判断list 是否有NULL 情况
  2. 同时遍历 两个新链表
  3. 如果节点地址相同,返回
  4. 如果不相同继续遍历
  5. 遍历结束返回NULL
 1 #include <iostream> 2 #include <unordered_map> 3 using namespace std; 4  5 /** 6  * Definition for singly-linked list. 7  */ 8  struct ListNode { 9      int val;10      ListNode *next;11      ListNode(int x) : val(x), next(NULL) {}12  };13 14 /**15 class Solution {16 public:17     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {18         unordered_map<ListNode*,int> m;19         while(headA!=NULL){20             m[headA] = 1;21             headA=headA->next;22         }23         while(headB!=NULL){24             if(m[headB]==1) return headB;25             headB=headB->next;26         }27         return NULL;28     }29 };30 */31 class Solution{32 public:33     ListNode* getIntersectionNode(ListNode *headA,ListNode * headB)34     {35         ListNode * h1=headA;36         ListNode * h2=headB;37         if(headA==NULL||headB==NULL)    return NULL;38         bool flag1=true,flag2=true;39         while(headA!=NULL&&headB!=NULL){40             if(headA==headB)    return headA;41             headA=headA->next;42             headB=headB->next;43             if(headA==NULL&&flag1){ headA=h2;   flag1 =false;}44             if(headB==NULL&&flag2){ headB=h1;   flag2 =false;}45         }46         return NULL;47     }48 };49 50 int main()51 {52     ListNode head1(1);53     ListNode head2(2);54     ListNode node1(3);55     ListNode node2(4);56     head1.next = &node1;57     node1.next = &node2;58     head2.next = &node2;59     Solution sol;60     ListNode *ret = sol.getIntersectionNode(&head1,&head2);61     if(ret==NULL)   cout<<"NULL"<<endl;62     else    cout<<ret->val<<endl;63     return 0;64 }
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[LeetCode] Intersection of Two Linked Lists 两链表是否相交