160. Intersection of Two Linked Lists

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本文介绍了一种高效的方法来找到两个单链表开始相交的节点。通过巧妙地利用两个指针，可以在O(n)的时间复杂度和O(1)的空间复杂度下解决问题，即使链表中存在共享部分。

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

egin 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.

my。。 too long

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int n = 0, m = 0;
        ListNode *pp = headA;
        ListNode *qq = headB;
        while(pp){
            n++;
            pp = pp->next;
        }
        while(qq){
            m++;
            qq = qq->next;
        }
        int i = 0;
        ListNode *p = headA;
        ListNode *q = headB;
        if(n < m){
            while(q){
                ++i;
                q = q->next;
                if(i == m-n)
                break;
                
            }
            
        }
        if(m < n){
            while(p){
                ++i;
                p = p->next;
                if(i == n-m)
                break;
                
            }
            
        }
        while(p){
            if(p == q)
            return p;
            p = p->next;
            q = q->next;
        }
        return NULL;
        
    }
};

Brilliant solution!

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    ListNode *cur1 = headA, *cur2 = headB;
    while(cur1 != cur2){
        cur1 = cur1?cur1->next:headB;
        cur2 = cur2?cur2->next:headA;
    }
    return cur1;
}