題目
描述
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node’s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node’s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
Example 3:
Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.
解法
c++ 保證兩個鏈表尾部長度一樣
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode *p1 = headA;
ListNode *p2 = headB;
ListNode *ret = NULL;
int len1 =0;
int len2 =0;
while(p1 != NULL){
len1++;
p1 = p1->next;
}
while(p2 != NULL){
len2++;
p2 = p2->next;
}
p1 = headA;
p2 = headB;
if (len1>len2)
{
int diff = len1-len2;
for (int i =0;i<diff;i++){
p1=p1->next;
}
}
else{
int diff = len2-len1;
for (int i=0 ;i<diff;i++){
p2=p2->next;
}
}
while(p1!=p2)
{
p1= p1->next;
p2=p2->next;
}
return p1;
}
};
python
class Solution(object):
def getIntersectionNode(self, headA, headB):
“”"
:type head1, head1: ListNode
:rtype: ListNode
“”"
if headA is None or headB is None:
return None
point_a,point_b = headA,headB
while point_a is not point_b:
point_a = point_a.next if point_a is not None else headB
point_b = point_b.next if point_b is not None else headA
return point_a