力扣面试题02.07-链表相交

链表相交

题目链接

解题思路：

1. 题目可以确定如果相交，那么相交的部分一定是在链表的结尾部分
2. 第一步求得两条链表的长度
3. 第二步长度做差，将长的那条链表与短的那条链表后部分对其
4. 第三步遍历后面的部分，如果当前节点相等，直接返回，否则返回null

/**
 * 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) {
        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;
        }
        curA = headA;
        curB = headB;
        if (lenA > lenB) {
            int temp = lenA - lenB;
            while (temp--) {
                curA = curA->next;
            }
            while (curA != NULL) {
                if (curA == curB) {
                    return curA;
                }
                curA = curA->next;
                curB = curB->next;
            }
        }else{
            int temp = lenB - lenA;
            while (temp--) {
                curB = curB->next;
            }
            while (curB != NULL) {
                if (curA == curB) {
                    return curA;
                }
                curA = curA->next;
                curB = curB->next;
            }
        }

        return NULL;
    }
};