题目

• 原题连接：160. 相交链表

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1- 思路

• 模式识别：相交链表 ——> 判断是否相交

思路

• 保证 headA 是最长的那个链表，之后对其开始依次遍历

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2- 实现

⭐160. 相交链表——题解思路

public class Solution {public ListNode getIntersectionNode(ListNode headA, ListNode headB) {// 求 headA 和 headB 长度int lenA = 0;int lenB = 0;ListNode curA = headA;ListNode curB = headB;while(curA!=null){lenA++;curA = curA.next;}while(curB!=null){lenB++;curB = curB.next;}if(lenB>lenA){ListNode tmp = headB;headB = headA;headA = tmp;int lenTmp = lenA;lenA = lenB;lenB = lenTmp;}curA = headA;curB = headB;for(int i = 0 ; i < lenA-lenB ;i++){curA = curA.next;}while(curA!=null  && curB!=null){if(curA == curB){return curA;}curA = curA.next;curB = curB.next;}return null;}
}

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3- ACM 实现

public class intersectLink {static class ListNode{int val;ListNode next;ListNode(){}ListNode(int x){val =x;}}public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {// 求 headA 和 headB 长度int lenA = 0;int lenB = 0;ListNode curA = headA;ListNode curB = headB;while(curA!=null){lenA++;curA = curA.next;}while(curB!=null){lenB++;curB = curB.next;}if(lenB>lenA){ListNode tmp = headB;headB = headA;headA = tmp;int lenTmp = lenA;lenA = lenB;lenB = lenTmp;}curA = headA;curB = headB;for(int i = 0 ; i < lenA-lenB ;i++){curA = curA.next;}while(curA!=null  && curB!=null){if(curA == curB){return curA;}curA = curA.next;curB = curB.next;}return null;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println("输入链表A和B的长度");int lenA = sc.nextInt();int lenB = sc.nextInt();System.out.println("输入链表A元素");ListNode headA = new ListNode(-1);ListNode curA = headA;for(int i = 0 ; i < lenA; i++){curA.next = new ListNode(sc.nextInt());curA = curA.next;}System.out.println("输入链表B的元素");ListNode headB = new ListNode(-1);ListNode curB = headB;for(int i = 0 ; i < lenB;i++){curB.next = new ListNode(sc.nextInt());curB = curB.next;}System.out.println("输入相交链表的长度");int intersectLen = sc.nextInt();ListNode intersectHead = new ListNode(-1);ListNode curIntersect = intersectHead;System.out.println("输入相交链表");for(int i = 0 ; i < intersectLen;i++){curIntersect.next = new ListNode(sc.nextInt());curIntersect = curIntersect.next;}if (intersectLen > 0) {curA.next = intersectHead.next;curB.next = intersectHead.next;}ListNode forRes = getIntersectionNode(headA,headB);System.out.println(forRes.val);}}

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