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题目：
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:
null.思路：求一个单链表交集的首结点
方法一：找出两个链表的长度差n，长链表先走n步，然后同时移动，判断有没有相同结点
方法二：两个链表同时移动，链表到尾部后，跳到链表头部，相遇点即为所求
代码：
方法一：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { 12 //方法一 13 ListNode *p = headA; 14 ListNode *q = headB; 15 if(!p || !q) return NULL; 16 while(p && q && p != q) 17 { 18 p = p->next; 19 q = q->next; 20 if(p == q) 21 return p; 22 if(!p) 23 p = headB; 24 if(!q) 25 q = headA; 26 } 27 return p; 28 } 29 };
方法二：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { 12 //方法二 13 if (headA == NULL || headB == NULL) return NULL; 14 ListNode *p = headA; 15 ListNode *q = headB; 16 int countA = 1, countB = 1; 17 while (p->next) 18 { 19 p = p->next; 20 ++countA; 21 } 22 while (q->next) 23 { 24 q = q->next; 25 ++countB; 26 } 27 if (p != q) return NULL; 28 if (countA > countB) 29 { 30 for (int i = 0; i < countA - countB; ++i) 31 headA = headA->next; 32 } 33 else if (countB > countA) 34 { 35 for (int i = 0; i < countB - countA; ++i) 36 headB = headB->next; 37 } 38 while (headA != headB) 39 { 40 headA = headA->next; 41 headB = headB->next; 42 } 43 return headA; 44 45 } 46 };
[LeetCode160]Intersection of Two Linked Lists
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原文地址：http://www.cnblogs.com/zhangbaochong/p/5186026.html
