LeetCode題解——Intersection of Two Linked Lists

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:
    //想法一：length1 - length2 = dis;兩個指針，長的連結清單先走dis步，然後一起走，指導遇見相同的節點。
    //特殊情況：空連結清單，連結清單在頭結點相交，連結清單在尾節點相交
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA || !headB) return NULL;
        if(headA==headB) return headA;
        // 周遊兩個連結清單，得到兩個連結清單的長度
        ListNode *pA  = headA;
        ListNode *pB  = headB;
        int lengthA=0,lengthB=0;
        while(pA){//注意條件判斷
            lengthA++;
            pA = pA->next;
        }
        while(pB){
            lengthB++;
            pB = pB->next;
        }
        ListNode *longlist = (lengthA>=lengthB)?headA:headB;
        ListNode *shortlist = (lengthA>=lengthB)?headB:headA;
        int dis = abs(lengthA-lengthB);
        for(int i=0;i<dis;i++){
            longlist = longlist->next;
        }
        while(longlist){
            if(longlist==shortlist) return longlist;
            longlist = longlist->next;
            shortlist = shortlist->next;
        }
        return NULL;
    }
};