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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.找到两个单链表相交的点。此处假设单链表不存在环。
思路：首先获得两个单链表的长度，得到长度的差值n，然后将指向长链表的指针A先向后移动n位，之后指针A同指向短链表的指针B同时每次移动一位，当A与B指向同一节点时，该节点就是相交的结点。
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 if(headA == NULL || headB == NULL) 13 { 14 return NULL; 15 } 16 ListNode * a = headA; 17 ListNode * b = headB; 18 int n = 1; 19 int m = 1; 20 while(headA->next != NULL) 21 { 22 headA = headA->next; 23 n++; 24 } 25 while(headB->next != NULL) 26 { 27 headB = headB->next; 28 m++; 29 } 30 if(headA != headB) 31 { 32 return NULL; 33 } 34 if(n > m) 35 { 36 for(int i = 0; i < n-m; i++) 37 { 38 a = a->next; 39 } 40 while(a) 41 { 42 if(a == b) 43 { 44 return a; 45 } 46 else 47 { 48 a = a->next; 49 b = b->next; 50 } 51 } 52 } 53 else 54 { 55 for(int i = 0; i < m-n; i++) 56 { 57 b = b->next; 58 } 59 while(b) 60 { 61 if(a == b) 62 { 63 return a; 64 } 65 else 66 { 67 a = a->next; 68 b = b->next; 69 } 70 } 71 } 72 return NULL; 73 } 74 };
拓展：
http://www.cppblog.com/humanchao/archive/2015/03/22/47357.html
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原文地址：http://www.cnblogs.com/shellfishsplace/p/5948333.html
