[LeetCode] Intersection of Two Linked Lists 两链表是否相交
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.
- 遍历list1 ,将其节点存在hash_table
- 遍历list2,如果已经在hash_table中,那么存在
利用hash_table 可以提升时间到O(n+m),可是空间变O(n)了
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
unordered_map<ListNode*,int> m;
while(headA!=NULL){
m[headA] = ;
headA=headA->next;
}
while(headB!=NULL){
if(m[headB]==) return headB;
headB=headB->next;
}
return NULL;
}
};
- 判断list 是否有NULL 情况
- 同时遍历 两个新链表
- 如果节点地址相同,返回
- 如果不相同继续遍历
- 遍历结束返回NULL
#include <iostream>
#include <unordered_map>
using namespace std; /**
* 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) {
unordered_map<ListNode*,int> m;
while(headA!=NULL){
m[headA] = 1;
headA=headA->next;
}
while(headB!=NULL){
if(m[headB]==1) return headB;
headB=headB->next;
}
return NULL;
}
};
*/
class Solution{
public:
ListNode* getIntersectionNode(ListNode *headA,ListNode * headB)
{
ListNode * h1=headA;
ListNode * h2=headB;
if(headA==NULL||headB==NULL) return NULL;
bool flag1=true,flag2=true;
while(headA!=NULL&&headB!=NULL){
if(headA==headB) return headA;
headA=headA->next;
headB=headB->next;
if(headA==NULL&&flag1){ headA=h2; flag1 =false;}
if(headB==NULL&&flag2){ headB=h1; flag2 =false;}
}
return NULL;
}
}; int main()
{
ListNode head1();
ListNode head2();
ListNode node1();
ListNode node2();
head1.next = &node1;
node1.next = &node2;
head2.next = &node2;
Solution sol;
ListNode *ret = sol.getIntersectionNode(&head1,&head2);
if(ret==NULL) cout<<"NULL"<<endl;
else cout<<ret->val<<endl;
return ;
}
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