Intersection of Two Linked Lists1、题目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 → b3begin 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.
2、分析题目大意：给两个链表，找出它们交集的那个节点，要求时间复杂度O(n)，空间复杂度O(1)。 有以下几种思路： （1） 暴力破解 ，遍历链表A的所有节点，并且对于每个节点，都与链表B中的所有节点比较，退出条件是在B中找到第一个相等的节点。时间复杂度O(lengthA*lengthB)，空间复杂度O(1)。 （2） 哈希表 。遍历链表A，并且将节点存储到哈希表中。接着遍历链表B，对于B中的每个节点，查找哈希表，如果在哈希表中找到了，说明是交集开始的那个节点。时间复杂度O(lengthA+lengthB)，空间复杂度O(lengthA)或O(lengthB)。 （3） 双指针法 ，指针pa、pb分别指向链表A和B的首节点。 遍历链表A，记录其长度lengthA，遍历链表B，记录其长度lengthB。 因为两个链表的长度可能不相同，比如题目所给的case，lengthA=5，lengthB=6，则作差得到 lengthB- lengthA=1，将指针pb从链表B的首节点开始走1步，即指向了第二个节点，pa指向链表A首节点，然后它们同时走，每次都走一步，当它们相等时，就是交集的节点。 时间复杂度O(lengthA+lengthB)，空间复杂度O(1)。双指针法的代码如下： 3、代码ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode *pa=headA,*pb=headB;
int lengthA=0,lengthB=0;
while(pa) {pa=pa->next;lengthA++;}
while(pb) {pb=pb->next;lengthB++;}
if(lengthA<=lengthB){
int n=lengthB-lengthA;
pa=headA;pb=headB;
while(n) {pb=pb->next;n--;}
}else{
int n=lengthA-lengthB;
pa=headA;pb=headB;
while(n) {pa=pa->next;n--;}
}
while(pa!=pb){
pa=pa->next;
pb=pb->next;
}
return pa;
}4、题目作者的解答（英文）作者双指针法花费的时间比我的方法更少
There are many solutions to this problem: - Brute-force solution (O(mn) running time, O(1) memory):
For each node a i in list A, traverse the entire list B and check if any node in list B coincides with a i . - Hashset solution (O(n+m) running time, O(n) or O(m) memory):
Traverse list A and store the address / reference to each node in a hash set. Then check every node b i in list B: if b i appears in the hash set, then b i is the intersection node. - Two pointer solution (O(n+m) running time, O(1) memory):
- Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
- When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
- If at any point pA meets pB, then pA/pB is the intersection node.
- To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
- If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
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