[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.
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?
- 遍历list1 ,将其节点存在hash_table
- 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 };
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:
- 判断list 是否有NULL 情况
- 同时遍历 两个新链表
- 如果节点地址相同,返回
- 如果不相同继续遍历
- 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 }