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, returnnull.
• 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.

Credits:
Special thanks to@stellarifor adding this problem and creating all test cases.

求两个链表的交点，要求Time: O(n), Space: O(1)

解法1：交点最早可能出现在短链表的第一个节点，后面的节点两个链表一样。所以，长链表的比短链表开始多出的那些就没用。求出两个链表的长度差值，把较长的链表向后移动这个差值，变成一样长。然后在一个一个的比较。

解法2:虽然题中强调链表不存在环，但可以用环的思想来做，让两条链表分别从各自的开头开始往后遍历，当其中一条遍历到末尾时，跳到另一个条链表的开头继续遍历。两个指针最终会相等，而且只有两种情况，一种情况是在交点处相遇，另一种情况是在各自的末尾的空节点处相等。因为两个指针走过的路程相同，是两个链表的长度之和，所以一定会相等。

Java：

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        int lenA = getLength(headA), lenB = getLength(headB);
        if (lenA > lenB) {
            for (int i = 0; i < lenA - lenB; ++i) headA = headA.next;
        } else {
            for (int i = 0; i < lenB - lenA; ++i) headB = headB.next;
        }
        while (headA != null && headB != null && headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }
        return (headA != null && headB != null) ? headA : null;
    }
    public int getLength(ListNode head) {
        int cnt = 0;
        while (head != null) {
            ++cnt;
            head = head.next;
        }
        return cnt;
    }
}

Java:

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        ListNode a = headA, b = headB;
        while (a != b) {
            a = (a != null) ? a.next : headB;
            b = (b != null) ? b.next : headA;
        }
        return a;
    }
}　

Python:

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
        begin, tailA, tailB = None, None, None
        
        # a->c->b->c
        # b->c->a->c
        while curA and curB:
            if curA == curB:
                begin = curA
                break
                
            if curA.next:
                curA = curA.next
            elif tailA is None:
                tailA = curA
                curA = headB
            else:
                break
            
            if curB.next:
                curB = curB.next
            elif tailB is None:
                tailB = curB
                curB = headA
            else:
                break
        
        return begin　　

C++：

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (!headA || !headB) return NULL;
        int lenA = getLength(headA), lenB = getLength(headB);
        if (lenA < lenB) {
            for (int i = 0; i < lenB - lenA; ++i) headB = headB->next;
        } else {
            for (int i = 0; i < lenA - lenB; ++i) headA = headA->next;
        }
        while (headA && headB && headA != headB) {
            headA = headA->next;
            headB = headB->next;
        }
        return (headA && headB) ? headA : NULL;
    }
    int getLength(ListNode* head) {
        int cnt = 0;
        while (head) {
            ++cnt;
            head = head->next;
        }
        return cnt;
    }
};

C++:

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

All LeetCode Questions List 题目汇总