[leetcode]Intersection of Two Linked Lists

Intersection of Two Linked Lists

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.
解题思路：先判断两个链表的长度，然后将长的链表长出的位置先移动

/**
* 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) {
ListNode *ptrA = headA;
ListNode *ptrB = headB;
ListNode *ptr = NULL;
//两个指针同时移动
while(ptrA && ptrB){
ptrA = ptrA->next;
ptrB = ptrB->next;
}
//如果一个为空则表示该链表较长，再用两个指针移动出长出的位置
if(ptrA == NULL && ptrB){
ptrA = headA;
while(ptrB && headB){
ptrB = ptrB->next;
headB = headB->next;
}
ptrB = headB;
}
if(ptrB == NULL && ptrA){
ptrB = headB;
while(ptrA){
ptrA = ptrA->next;
headA = headA->next;
}
ptrA = headA;
}
if(ptrA == NULL && ptrB == NULL){ //两个链表一样长
ptrA = headA;
ptrB = headB;
}
while(ptrA && ptrB){
if(ptrA == ptrB) return ptrA;
ptrA = ptrA->next;
ptrB = ptrB->next;
}

return NULL;
}
};