LeetCode OJ 160. Intersection of Two Linked Lists

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Description

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.

Note:

• 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.

方法一

比较简单直观的一个思路：计算链表长度，然后长的链表指针先走，直到剩下长度等于短的链表长度，然后两个链表指针一起移动，直到两个指针相遇，就返回。
两个指针遍历两次链表，算两个链表长度为m，n，那么时间复杂度为O(m+n)，空间复杂度为O(1)，只用了几个变量用于求链表长度并保存下来。

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    int getLen(ListNode *head){        int l = 0;        while(head){            head = head->next;            l ++;        }        return l;    }    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {        int lenA, lenB;        lenA = getLen(headA);        lenB = getLen(headB);        if(lenA < lenB){            int temp = lenA;            lenA = lenB;            lenB = temp;            ListNode *node = headA;            headA = headB;            headB = node;        }        while(lenA > lenB)        {            headA = headA->next;            lenA --;        }        while(headA !=NULL && headA != headB){            headA = headA->next;            headB = headB->next;        }        return headA;    }};

方法二

两个指针p,q分别指向两个链表表头，然后分别每次移动一个节点来遍历两个链表。如果链表A到了尾节点，则指向链表B的头节点；同理，如果链表B到了尾节点也指向A的表头。这样p,q走过的路程最坏的情况就是两个链表的长度(m+n)，如果p,q相遇的话，就是那个相遇的节点。所以时间复杂度为O(m+n)，空间复杂度为O(1)。

代码

个人github代码链接

/** * 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 *p,*q;        p = headA;        q = headB;        if(p == NULL || q == NULL)   return NULL;        while(p && q && p != q){            p = p->next;            q = q->next;            if(p == q) return p;            if(p == NULL) p = headB;            if(q == NULL) q = headA;        }        return p;    }};