160. Intersection of Two Linked Lists(Linked List-Easy)

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.

题目：写一个程序找出两个单链表的交叉节点。

思路：单链表A和单链表B，交叉点后的部分是一样的，也就是说长度是一样的，如上所示：c1 → c2 → c3。所以，将单链表A和单链表B相差的部分去掉，依次对应比较等长的部分即可。在计算两个链表的长度之后，比较两个链表的尾节点是否一样，如果不一样说明没有交叉节点，返回NULL。

Language : c

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
struct ListNode *curA = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode *curB = (struct ListNode*)malloc(sizeof(struct ListNode));
curA = headA;
curB = headB;
int length_a = 1;
int length_b = 1;
int i = 0;
if(curA == NULL || curB == NULL){
return NULL;
}
while(curA->next != NULL){
curA = curA->next;
length_a++;
}
while(curB->next != NULL){
curB = curB->next;
length_b++;
}
if(curA != curB){
return NULL;
}
curA = headA;
curB = headB;
if(length_a > length_b){
for(i; i < length_a-length_b; i++){
curA = curA->next;
}
i = 0;
}
else if(length_a < length_b){
for(i; i < length_b-length_a; i++){
curB = curB->next;
}
i = 0;
}
while(curA != curB){
curA = curA->next;
curB = curB->next;
}
return curA;
}

Language : cpp

/**
* 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 *curA, *curB;
curA = headA;
curB = headB;
if(curA == NULL || curB == NULL){
return NULL;
}
int length_a = getLength(curA);
int length_b = getLength(curB);
if(length_a > length_b){
for(int i=0; i < length_a-length_b; i++){
curA = curA->next;
}
}
else if(length_a < length_b){
for(int i=0; i < length_b-length_a; i++){
curB = curB->next;
}
}
while(curA != curB){
curA = curA->next;
curB = curB->next;
}
return curA;
}
private:
int getLength(ListNode *head){
int length = 1;
while(head->next != NULL){
head = head->next;
length++;
}
return length;
}
};

Language：python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
if headA is None or headB is None:
return None

pa = headA # 2 pointers
pb = headB

while pa is not pb:
# pa先遍历headA，然后再遍历headB
# pb先遍历headB，然后再遍历headA
pa = headB if pa is None else pa.next
pb = headA if pb is None else pb.next

return pa # 只有两种方式结束循环，一种是pa和pb所指相同，另一种是headA和headB都已经遍历完仍然没有找到。