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160. intersection of two linked lists 解法

admin 11月 12, 2020 0

160. Intersection of Two Linked Lists

Difficulty: Easy

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:

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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:

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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:

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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.

解法

  这道题看到了一个比较新并且代码简洁的解法，所以写了一篇记录一下。

法一

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 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *(struct ListNode *headA, struct ListNode *headB) {
    int i, j = 0;
    struct ListNode *pA = headA;
    struct ListNode *pB = headB;
    while(pA != NULL){
        pA = pA->next;
        i++;
    }
    while(pB != NULL){
        pB = pB->next;
        j++;
    }
    if(i < j){
        int k = j - i;
        while(k--) headB = headB->next;
    }
    else{
        int k = i - j;
        while(k--) headA = headA->next;
    }
    while(headA != headB){
        headA = headA->next;
        headB = headB->next;
    }
    return headA;
}

法二

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 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *pA = headA;
    struct ListNode *pB = headB;
    while(pA != pB){
        if(pA == NULL) 
            pA = headB;
        else
            pA = pA->next;
        if(pB == NULL)
            pB = headA;
        else
            pB = pB->next;
    }
    return pA;    
}
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