LeetCode 160. Intersection of Two Linked Lists

这个题目较容易，我的想法是先比较两个链表的长度，我们要找到的交叉点是不可能为长的链表的前面超长的部分的，

 

所以要比较的就是两者之间较短的部分

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

  /**
   * 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) {
          size_t lengthA=0, lengthB=0,minus=0;
          ListNode *pA=headA, *pB=headB,*IntersectionNode=NULL;
          while (pA != NULL)
          {
              ++lengthA;
              pA = pA->next;
          }
          while (pB != NULL)
          {
              ++lengthB;
              pB = pB->next;
          }
          if (lengthA > lengthB)
          {
              pA = headA;
              pB = headB;
              minus = lengthA - lengthB;
              while (minus--)
              {
                  pA = pA->next;
              }
              while (pA != NULL&&pB != NULL)
              {
                  if (pA->val == pB->val&&IntersectionNode==NULL)
                      IntersectionNode = pA;
                  if (pA->val != pB->val)
                      IntersectionNode = NULL;
                  pA = pA->next;
                  pB = pB->next;
              }
              return IntersectionNode;
          }
          else
          {
              pA = headA;
              pB = headB;
              minus = lengthB - lengthA;
              while (minus--)
              {
                  pB = pB->next;
              }
              while (pA != NULL&&pB != NULL)
              {
                  if (pA->val == pB->val&&IntersectionNode == NULL)
                      IntersectionNode = pA;
                  if (pA->val != pB->val)
                      IntersectionNode = NULL;
                  pA = pA->next;
                  pB = pB->next;
              }
              return IntersectionNode;
          }
      }
  };

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