[LeetCode] 160. Intersection of Two Linked Lists 求两个链表的交集

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. 

Credits:
Special thanks to @stellari for 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: 双指针，用两个指针pA和pB分别指向链表A和B。然后让它们分别遍历整个链表，每步一个节点。当pA到达链表末尾时，让它指向B的头节点（没错，是B）；类似的当pB到达链表末尾时，重新指向A的头节点。如果pA在某一点与pB相遇，则pA/pB就是交集开始的节点。

Java： Solution 1

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: Solution 2

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 Solution:
    # @param two ListNodes
    # @return the intersected ListNode
    def getIntersectionNode(self, headA, headB):
        if headA is None or headB is None:
            return None

        pa = headA # 2 pointers
        pb = headB

        while pa is not pb:
            # if either pointer hits the end, switch head and continue the second traversal, 
            # if not hit the end, just move on to next
            pa = headB if pa is None else pa.next
            pb = headA if pb is None else pb.next

        return pa # only 2 ways to get out of the loop, they meet or the both hit the end=None

# the idea is if you switch head, the possible difference between length would be countered. 
# On the second traversal, they either hit or miss. 
# if they meet, pa or pb would be the node we are looking for, 
# if they didn't meet, they will hit the end at the same iteration, pa == pb == None, return either one of them is the same,None

Python: wo

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        if not headA or not headB:
            return None

        a, b = headA, headB
        while a != b:
            a = a.next if a else headB
            b = b.next if b else headA
            
        return a 　

Python: Solution 1

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        lenA = self.getListLen(headA)
        lenB = self.getListLen(headB)
        if lenA > lenB:
            for i in range(lenA - lenB):
                headA = headA.next
        elif lenA < lenB:
            for i in range(lenB - lenA):
                headB = headB.next
        while headA != headB:
            headA, headB = headA.next, headB.next
        return headA

    def getListLen(self, head):
        length = 0
        while head:
            length += 1
            head = head.next
        return length　

Python: Solution 2

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;
    }
};

　

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