160. 相交链表

目录

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

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：常规

• 先求ab长度，长的先走差值，再一起走，相同则返回

var getIntersectionNode = function(headA, headB) {
    if(headA === null || headB === null) return null;

    let pA = headA;
    let pB = headB;
    let lengthA = 0;
    let lengthB = 0;
    //求链表长度
    while(pA.next!==null){
        lengthA++;
        pA = pA.next;
    }
    while(pB.next!==null){
        lengthB++;
        pB = pB.next;
    }
    //上一步求完长度指针在链表尾，重新放回头
    pA = headA;
    pB = headB;
    //较长的链表先走差值
    if(lengthA>lengthB){
        let n = lengthA-lengthB;
        while(n!=0){
            pA = pA.next;
            n--;
        }
    }else if(lengthA<lengthB){
        let n = lengthB-lengthA;
        while(n!=0){
            pB = pB.next;
            n--;
        }
    }
    //一样长后，一起走，遇到相同返回
    while(pA!==null && pB!==null){
        if(pA===pB){
            return pA;
        }
        pA=pA.next;
        pB=pB.next;
    }
    return null;//走完都没相交
};

题解：技巧

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
};