LeetCode—Intersection of Two Linked Lists 找兩個鏈表的相交位置，讓長的鏈表先走一段

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. 只能夠是O（n）的時間復雜度，這裡的做法是首先遍歷兩個鏈表知道長度，然後讓長的鏈表先走一段，然後再進行兩兩比較，時間復雜度O（lengthA+lengthB+maxLength）就是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) {
          if(headA == NULL || headB == NULL)
          {
              return NULL;
          }
          int lengthA = 1;
          int lengthB = 1;
          ListNode *Afirst = headA;
          ListNode *Bfirst = headB;
          while(Afirst->next)
          {
              lengthA++;
              Afirst = Afirst->next;
          }
          while(Bfirst->next)
          {
              lengthB++;
              Bfirst = Bfirst->next;
          }
          if(Afirst != Bfirst)
          {
              return NULL;
          }
          int length = lengthA;//<隨便初始化一個
          if(lengthA > lengthB)
          {
              int diff = lengthA-lengthB;
              length = lengthB;
              while(diff--)
              {
                  headA = headA->next;
              }
          }
          if(lengthA < lengthB)
          {
              length = lengthA;
              int diff = lengthB-lengthA;
              while(diff--)
              {
                  headB = headB->next;
              }
          }
          while(length--)
          {
              if(headA == headB)
              {
                  return headA;
              }
              headA = headA->next;
              headB = headB->next;
          }
          return NULL;
          
      }
  };