思路:
“移动快慢指针”的思想:先让长链表的指针(假定为A)走 lenA - lenB
的距离,再两个指针一起走,碰到相同的点就有intersection。
时间复杂度O(N),空间复杂度O(1)。
/**
* 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 = 0, lenB = 0;
ListNode *p1 = headA, *p2 = headB;
while(p1 != NULL) {
lenA++;
p1 = p1->next;
}
while(p2 != NULL) {
lenB++;
p2 = p2->next;
}
int dis = 0;
p1 = headA, p2 = headB;
if(lenA >= lenB) {
dis = lenA - lenB;
while(dis != 0) {
p1 = p1->next;
dis--;
}
}else {
dis = lenB - lenA;
while(dis != 0) {
p2 = p2->next;
dis--;
}
}
while(p1 != NULL && p2 != NULL) {
if(p1 == p2) {
return p1;
}else {
p1 = p1->next;
p2 = p2->next;
}
}
return NULL;
}
};