160. 相交链表

目录

• 题目
• python
  • 题解：双指针
• javascript
  • 题解：双指针

题目

python

题解：双指针

• 思路：计算两条链表的长度，找到长度差，让长的链表多走差的值，返回第一个相等的元素

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        count1,count2=0,0
        pa=headA
        pb=headB
        while headA:# 计算链表 A 的长度
            headA=headA.next
            count1+=1
        while headB:# 计算链表 B 的长度
            headB=headB.next
            count2+=1
        if count1>count2:
            count=count1-count2# 计算长度差
            #让A链先走count步
            while count!=0:
                pa=pa.next
                count-=1
        elif count1<count2:
            count=count2-count1# 计算长度差
            #让B链先走count步
            while count!=0:
                pb=pb.next
                count-=1
        while pa and pb:# 同时遍历链表 A 和链表 B，找到第一个相等的节点
            if pa==pb:
                return pa
            pa=pa.next
            pb=pb.next
        return None# 如果没有交点，返回 None      

javascript

题解：双指针

var getIntersectionNode = function(headA, headB) {
    if(headA === null || headB === null){
        return null
    }
    let pA = headA, pB = headB
    while(pA!==pB){
        pA = pA === null?headB:pA.next
        pB = pB ===null?headA:pB.next
    }
    return pA
};