-
Notifications
You must be signed in to change notification settings - Fork 0
/
linked-list-intersection.cpp
52 lines (44 loc) · 1.42 KB
/
linked-list-intersection.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
/* LC 160
* 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.
*/
#include<iostream>
#include<cmath>
using namespace std;
int count(ListNode* h) {
int c = 0;
while(h) {
c++;h = h->next;
}
return c;
}
void advance(ListNode* &l, int d) {
while(d--) {
l = l->next;
}
}
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
int c1 = count(headA);
int c2 = count(headB);
advance(c1>c2?headA:headB,abs(c1-c2));
while(headA && headB && headA!=headB) {
headA = headA->next;
headB = headB->next;
}
return headA;
}
/* Tested.
* Time complexity: O(n+m) where n i& m are the number of nodes of the 2 lists.
* Space complexity:O(1)
*/