[LeetCode] Intersection of Two Linked Lists

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.

解题思路：

1、最naive的办法，就是最每个a的指针，在b里面找，是否与之相同，若找到，则返回，否则返回NULL。这个办法时间复杂度为O(n*m)。不符合要求。

2、可以考虑用set来存储a的每个指针，然后针对每个b的指针，在set中查找。因为set中的查找可以看成是O(1)的，因此时间复杂度为O(n)。但是空间复杂度为O(n)，不符合要求。

3、我的办法是，先分别扫描一遍a、b，记录a、b的长度为aLen，bLen。然后从头分别开始扫描，每扫描一个aLen--或bLen--，然后a或b向后移一个。具体是a移呢还是b移呢？这要看aLen大还是bLen大。具体代码如下，代码具有自解释：

/**
 * 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) {
        int aLen = 0, bLen = 0;
        ListNode *p, *q;
        p=headA;
        while(p!=NULL){
            aLen++;
            p=p->next;
        }
        p=headB;
        while(p!=NULL){
            bLen++;
            p=p->next;
        }
        ListNode* result=NULL;
        p=headA;
        q=headB;
        while(p!=NULL&&q!=NULL){
            if(p==q){
                result=p;
                break;
            }else{
                if(aLen>bLen){
                    p=p->next;
                    aLen--;
                }else if(aLen<bLen){
                    q=q->next;
                    bLen--;
                }else{
                    p=p->next;
                    q=q->next;
                    aLen--; //这里减不减都无所谓
                    bLen--;
                }
            }
        }
        return result;
    }
};

/**
 * 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, *pB=headB;
        while(pA!=NULL&&pB!=NULL){
            if(pA==pB){
                return pA;
            }
            pA=pA->next;
            pB=pB->next;
        }
        if(pA==NULL){   //B比A长
            pA=headB;
        }else{
            pB=headA;
        }
        while(pA!=NULL&&pB!=NULL){
            pA=pA->next;
            pB=pB->next;
        }
        if(pA==NULL){   //与之前第20步相反
            pA=headB;
        }else{
            pB=headA;
        }
        while(pA!=NULL&&pB!=NULL){
            if(pA==pB){
                return pA;
            }
            pA=pA->next;
            pB=pB->next;
        }
        return NULL;
    }
};

[LeetCode] Intersection of Two Linked Lists

原文：http://blog.csdn.net/kangrydotnet/article/details/45010357