leetcode-160-Intersection of Two Linked Lists

/**
 * 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 nA = 0,nB = 0,n;// 链表长度
        ListNode* p;
        
        p = headA; //headA的长度nA
        while (p) {
            nA++;
            p = p->next;
        }
        p = headB; //headB的长度nB
        while (p) {
            nB++;
            p = p->next;
        }
    
        if (nA < nB) {
            n = nB - nA;
            while (n--) headB = headB->next;
        } 
        else if (nA > nB) {
            n = nA - nB;
            while (n--) headA = headA->next;
        }
        
        while (headA && headB) {
            if (headA == headB) return headA;
            headA = headA->next;
            headB = headB->next;
        }
        return NULL;
    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param two ListNodes
    # @return the intersected ListNode
    def getIntersectionNode(self, headA, headB):
        nA = nB = 0
        p = headA
        while (p is not None) :
            nA = nA + 1
            p = p.next
        p = headB
        while (p is not None) :
            nB = nB + 1
            p = p.next
        
        if (nA < nB) :
            n = nB - nA
            while (n) :
                headB = headB.next
                n = n - 1
        elif (nA > nB) :
            n = nA - nB
            while (n) :
                headA = headA.next
                n = n - 1
        
        while (headA  and headB ) :
            if (headA == headB):
                return headA
            headA = headA.next
            headB = headB.next
        
        return None

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