剑指offer | 两个链表的第一个公共节点 | 16

目录

• 思路分析

思路分析

这个题目只有两种情况,如下图

cpp

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *findFirstCommonNode(ListNode *headA, ListNode *headB) {
        auto p=headA,q=headB;
        while(p!=q){
            p= p ? p->next : headB;
            q= q ? q->next : headA;
        }
        return q;
    }
};

python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        p = headA
        q = headB
        while p!=q: # 最后一定会相等
            p = p.next if p else headB
            q = q.next if q else headA
        return q