編寫一個程式，找到兩個單鏈表相交的起始節點。

例如，下面的兩個連結串列：

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

在節點 c1 開始相交。

注意：
如果兩個連結串列沒有交點，返回 null.
在返回結果後，兩個連結串列仍須保持原有的結構。
可假定整個連結串列結構中沒有迴圈。
程式儘量滿足 O(n) 時間複雜度，且僅用 O(1) 記憶體。

這個題如果使用暴力迴圈的話複雜度就會很高，可能會達到O( $n^{}$

/**
 * 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) {
        ListNode *pA = headA;
        ListNode *pB = headB;
        int lenA = 0, lenB = 0; 
        while(pA!=NULL || pB!=NULL){
            if (pA!=NULL){
                lenA++;
                pA = pA->
 
next;
            } 
            if (pB!=NULL){
                lenB++;
                pB = pB->next;
            } 
        }
        pA = headA;
        pB = headB;
        if (lenA>lenB){
            int diff = lenA - lenB;
            while(diff--)
                pA = pA->next;    
        }
        if(lenB>lenA){
            int diff = lenB - lenA;
            while(diff--)
                pB = pB->next;    
        }
        while(pA!=pB){
            pA = pA->next;
            pB = pB->next;
        }
        return pA;
    }
};

python3原始碼：

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

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        pA = headA
        pB = headB
        lenA = 0
        lenB = 0
        while pA!=None or pB!=None:
            if pA!=None:
                lenA += 1
                pA = pA.next
            if pB!=None:
                lenB += 1
                pB = pB.next
        pA = headA
        pB = headB
        if lenA > lenB:
            diff = lenA - lenB
            while diff:
                pA = pA.next
                diff -= 1
        if lenB > lenA:
            diff = lenB - lenA
            while diff:
                pB = pB.next
                diff -= 1
        while pA!=pB:
            pA = pA.next
            pB = pB.next
        return pA