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 → b3begin 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.
就是先找到相同长度的位置。再逐一对比。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { int n1 = getLength(headA); int n2 = getLength(headB); if(n1-n2>0){ for(int i =0;i<n1-n2;++i){ headA = headA->next; } }else{ for(int i =0;i<n2-n1;++i){ headB = headB->next; } } while(headA != headB){ headA = headA->next; headB = headB->next; } return headA; } private: int getLength(ListNode *head){ int res = 0; while(head){ head=head->next; res++; } return res; } };
********************round 2 ********two pointer *****
class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *curA = headA; ListNode *curB = headB; if (!headA || !headB) return NULL; while (curA != curB) { curA = curA? curA->next:headB; curB = curB? curB->next:headA; } return curA; } };
Two Pointers
- Maintain two pointers and initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
- When reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when reaches the end of a list, redirect it the head of A.
- If at any point meets , then / is the intersection node.
- To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), would reach the end of the merged list first, because traverses exactly 2 nodes less than does. By redirecting to head A, and to head B, we now ask to travel exactly 2 more nodes than would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
- If two lists have intersection, then their last nodes must be the same one. So when / reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
Complexity Analysis
- Time complexity : .
- Space complexity : .
Mistakes:
while(head){
这里 一开始写成了 !head
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