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:
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.

非常优雅的解决方案：

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) 
{
    ListNode *p1 = headA;
    ListNode *p2 = headB;

    if (p1 == NULL || p2 == NULL) return NULL;

    while (p1 != NULL && p2 != NULL && p1 != p2) 
    {
        p1 = p1->next;
        p2 = p2->next;

        //
        // Any time they collide or reach end together without colliding 
        // then return any one of the pointers.
        //
        if (p1 == p2) return p1;

        //
        // If one of them reaches the end earlier then reuse it 
        // by moving it to the beginning of other list.
        // Once both of them go through reassigning, 
        // they will be equidistant from the collision point.
        //
        if (p1 == NULL) p1 = headB;
        if (p2 == NULL) p2 = headA;
    }

    return p1;
    //上面似乎不需要，上面的路径应该都有返回值，要不要都ac
}

python解决方案：

# 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):
        curA,curB = headA,headB
        lenA,lenB = 0,0
        while curA is not None:
            lenA += 1
            curA = curA.next
        while curB is not None:
            lenB += 1
            curB = curB.next
        curA,curB = headA,headB
        if lenA > lenB:
            for i in range(lenA-lenB):
                curA = curA.next
        elif lenB > lenA:
            for i in range(lenB-lenA):
                curB = curB.next
        while curB != curA:
            curB = curB.next
            curA = curA.next
        return curA

/*
The solution is straightforward: maintaining two pointers in the lists under the constraint that both lists have the same number of nodes starting from the pointers. We need to calculate the length of each list though. So O(N) for time and O(1) for space.
*/