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.
思路:简单题 遍历长度 走差值 再同步走 答案里给了个转圈圈的思路 感觉没有自己的好 太乱了
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { if(headA == NULL || headB == NULL) { return NULL; } int lenA = 1; int lenB = 1; ListNode * pa = headA; ListNode * pb = headB; while(pa->next != NULL) { pa = pa->next; lenA++; } while(pb->next != NULL) { pb = pb->next; lenB++; } if(pa != pb) return NULL; pa = headA; pb = headB; while(lenA > lenB) { pa = pa->next; lenA--; } while(lenB > lenA) { pb = pb->next; lenB--; } while(pa != pb) { pa = pa->next; pb = pb->next; } }