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.
解題思路:
1、可以將A,B兩個鏈表看做兩部分,交叉前與交叉後。
2、交叉後的長度是一樣的,因此交叉前的長度差即爲總長度差。
3、只要去除這些長度差,距離交叉點就等距了。
4、爲了節省計算,在計算鏈表長度的時候,順便比較一下兩個鏈表的尾節點是否一樣,若不一樣,則不可能相交,直接可以返回NULL。
/**
* 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) {
if(headA == NULL || headB == NULL)
return NULL;
int lenA = 1;
int lenB = 1;
ListNode *tempA = headA;
ListNode *tempB = headB;
while(tempA->next != NULL){
tempA = tempA->next;
lenA++;
}
while(tempB->next != NULL){
tempB = tempB->next;
lenB++;
}
if(tempA != tempB)
return NULL;
if(lenA>lenB){
for(int i=0;i<lenA-lenB;i++){
headA = headA->next;
}
}
if(lenB>lenA){
for(int i=0;i<lenB-lenA;i++){
headB = headB->next;
}
}
while(headA != headB){
headA = headA->next;
headB = headB->next;
}
return headA;
}
};