160. Intersection of Two Linked Lists

Posted 2020-12-29 hozhangel

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:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

 

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.

---

 

My hints :

1、First walk two list saperately,and count the length of the two list:lenA,lenB
2、n = lenA - lenB,and let the point_A go n step first
3、then point_A,point_B go toghter,until point_A == point_B

My code:

 个人感觉自己的思路更好理解，时间复杂度为m+n。(#^.^#)

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        
        ListNode *p1 = headA;
        ListNode *p2 = headB;
        
        if (p1 == NULL || p2 == NULL) return NULL;
        
        int n1 = 0;
        int n2 = 0;
        int n = 0;
        while(p1 != NULL){
            n1++;
            p1 = p1->next;
        }
        while(p2 != NULL){
            n2++;
            p2 = p2->next;
        }
        
        p1 = headA;
        p2 = headB;
        if(n1 > n2){
            n = n1 - n2;
            while(n != 0){
                p1 = p1->next;
                n--;
            }
        }
        if(n1 < n2){
            n = n2 - n1;
            while(n != 0){
                p2 = p2->next;
                n--;
            }
        }
        
        while (p1 != NULL && p2 != NULL) {
            if (p1 == p2) return p1;
            p1 = p1->next;
            p2 = p2->next;
        }
        return NULL;
    }
};

 

discussion区还有更简单的算法My-accepted-simple-and-shortest-C++-code ，但是理解起来不是很好理解（个人感觉。。）