160. Intersection of Two Linked Lists (easy)

Q:

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. 

A:
就是先找到相同长度的位置。再逐一对比。

/**
 * 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 n1 = getLength(headA);
        int n2 = getLength(headB);        
        if(n1-n2>0){
            for(int i =0;i<n1-n2;++i){
                headA = headA->next;
            }
        }else{
            for(int i =0;i<n2-n1;++i){
                headB = headB->next;
            }
        }
        while(headA != headB){
            headA = headA->next;
            headB = headB->next;
        }
        return headA;
    }
private:
    int getLength(ListNode *head){
        int res = 0;
        while(head){
            head=head->next;
            res++;
        }
        return res;
    }
};

********************round 2 ********two pointer *****

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *curA = headA;
        ListNode *curB = headB;
        if (!headA || !headB) 
            return NULL;
        while (curA != curB)
        {
            curA = curA? curA->next:headB;
            curB = curB? curB->next:headA;
        }
        return curA;
    }
};

Two Pointers

• Maintain two pointers $pA$ and $pB$ initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
• When $pA$ reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when $pB$ reaches the end of a list, redirect it the head of A.
• If at any point $pA$ meets $pB$, then $pA$/$pB$ is the intersection node.
• To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), $pB$ would reach the end of the merged list first, because $pB$ traverses exactly 2 nodes less than $pA$ does. By redirecting $pB$ to head A, and $pA$ to head B, we now ask $pB$ to travel exactly 2 more nodes than $pA$ would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
• If two lists have intersection, then their last nodes must be the same one. So when $pA$/$pB$ reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.

Complexity Analysis

• Time complexity : $O(m+n)$.
• Space complexity : $O(1)$.

Mistakes:

        while(head){

这里 一开始写成了   !head