160. 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.

题解

返回两个链表第一个公共节点，没有返回NULL，不允许修改原链表，时间复杂度O(n)，空间复杂度O(1)，样例已保证无环。

解法1 尾对齐

具有公共部分的链表一定具有共同的尾部。先找出两个链表的长度，让长的链表先走，使两个链表靠尾部对齐，之后一起走直到指针相等。

/** * 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 || !headB)    return NULL;        int a = 0, b = 0;        ListNode* cur1 = headA, *cur2 = headB;        //记录长度        while(headA)    headA = headA->next, a++;        while(headB)    headB = headB->next, b++;        //为方便保证headA为短，headB为长        if(a > b){            swap(a, b);            swap(cur1, cur2);        }        //headB先走        for(int i = 0; i < b - a; i++)  cur2 = cur2->next;        //一起走        while(cur1 != cur2){            cur1 = cur1->next;            cur2 = cur2->next;        }        return cur1;    }};

解法2 你来我往

让两个指针都走一遍两个链表，则两个指针走得路程相同（a+b)，不用再计较谁长谁短，第一个相遇的点就是第一个公共的点。

class Solution {public:    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {        ListNode *cur1 = headA, *cur2 = headB;        while(cur1 != cur2){            cur1 = cur1?cur1->next:headB;            cur2 = cur2?cur2->next:headA;        }        return cur1;    }};