leetcode 160. Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

 

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

 

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

 

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

 

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.

Your code should preferably run in O(n) time and use only O(1) memory.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    // 时间复杂度O(n2), 空间复杂度O(1)
    // ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    //     ListNode *pa = headA, *pb;
    //     while(pa != NULL) {
    //         for (pb = headB; pb != NULL; pb = pb->next) {
    //             if (pb == pa)
    //                 return pa;
    //         }
    //         pa = pa->next;
    //     }
    //     return NULL;
    // }
    // 时间复杂度O(m + n), 空间复杂度O(m)
    // 哈希表，保存A的节点地址，找到B中第一个和A中节点一样的节点
    // ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    //     map<ListNode*, int> mp;
    //     if (headA == NULL || headB == NULL)
    //         return NULL;
    //     ListNode *pa = headA;
    //     while (pa != NULL) {
    //         mp[pa]++;
    //         pa = pa->next;
    //     }
    //     ListNode *pb = headB;
    //     while (pb != NULL) {
    //         if (mp.count(pb) != 0)
    //             return pb;
    //         pb = pb->next;
    //     }
    //     return NULL;
    // }
    // 时间复杂度O(k), 空间复杂度O(m + n)
    // 双栈，只要栈顶节点一样，一直往外弹出，找到最后一个相同的节点。
    // ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    //     stack<ListNode*> sa, sb;
    //     ListNode *pa = headA, *pb = headB;
    //     while (pa != NULL) {
    //         sa.push(pa);
    //         pa = pa->next;
    //     }
    //     while (pb != NULL) {
    //         sb.push(pb);
    //         pb = pb->next;
    //     }
    //     ListNode *res = NULL;
    //     while (!sa.empty() && !sb.empty() && sa.top() == sb.top()) {
    //         res = sa.top();
    //         sa.pop();
    //         sb.pop();
    //     }
    //     return res;
    // }
    // 时间复杂度O(m + n), 空间复杂度O(1)
    // ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    //     ListNode *pa = headA, *pb = headB;
    //     //把两个链表连接起来. pa: headA->headB, pb: headB->headA。通过这种方式会实现最后面公共节点的对齐，
    //     //有公共节点的时候，遇到就返回，无公共节点的时候，同时到达最后面的NULL，返回NULL
    //     while (pa != pb) {
    //         pa = pa != NULL ? pa->next : headB;
    //         pb = pb != NULL ? pb->next : headA;
    //     }
    //     return pa;
    // }
    // 时间复杂度：O(m + n), 空间复杂度O(1)
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *pa = headA, *pb = headB;
        int lena = 0;
        int lenb = 0;
        while (pa != NULL) {
            lena++;
            pa = pa->next;
        }
        while (pb != NULL) {
            lenb++;
            pb = pb->next;
        }
        pa = headA;
        pb = headB;
        if (lena > lenb) {
            int diff = lena - lenb;
            for (int i = 0; i < diff; i++) {
                pa = pa->next;
            }
        } else {
            int diff = lenb - lena;
            for (int i = 0; i < diff; i++) {
                pb = pb->next;
            }
        }
        while (pa != pb) {
            pa = pa->next;
            pb = pb->next;
        }
        return pa;
    }
};