leetcode刷题系列----模式2（Datastructure 链表）- 160：Intersection of Two Linked List 相交链表

leetcode刷题系列----模式2（Datastructure 链表）- 160：Intersection of Two Linked List 相交链表

Tips

• 更多题解请见本系列目录
• 此链表问题可以使用双指针解决，注意算法的复杂度为O(m+n),意味着2*（m+n）次遍历也属于该时间复杂度，不可狭隘的认为仅仅遍历m+n次链表节点。
• 双指针确实是一个很大的算法思想类。

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:
        pointA = headA
        pointB = headB
        if not pointA or not pointB: return None
        while pointA != pointB:
            if pointA: pointA = pointA.next
            else: pointA = headB
            if pointB: pointB = pointB.next
            else: pointB = headA
        return pointA

C++

/**
 * 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) {
        ListNode* pointA = headA;
        ListNode* pointB = headB;
        if (pointA==nullptr || pointB==nullptr) return nullptr;
        while (pointA!=pointB)
        {
            if(pointA!=nullptr) pointA=pointA->next;
            else pointA=headB;
            if(pointB!=nullptr) pointB=pointB->next;
            else pointB=headA;
        }
        return pointA;
        
    }
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
        ListNode pointA = headA;
        ListNode pointB = headB;
        if (pointA==null || pointB==null) return null;
        while (pointA!=pointB)
        {
            if(pointA!=null) pointA=pointA.next;
            else pointA=headB;
            if(pointB!=null) pointB=pointB.next;
            else pointB=headA;
        }
        return pointA;
    }
}

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode pointA = headA;
        ListNode pointB = headB;
        if (pointA==null || pointB==null) return null;
        while (pointA!=pointB)
        {
            if(pointA!=null) pointA=pointA.next;
            else pointA=headB;
            if(pointB!=null) pointB=pointB.next;
            else pointB=headA;
        }
        return pointA;        
    }
}