共谋节点两个单列表

题目：输入两个单链表。找出公共结点。

思路：若两个单链表有公共结点。其形状必然为“Y”型，也就是说公共结点后的全部结点都是同样的。

我们首先获得两个链表的长度。求得长度之差为n，再定义两个指针分别指向两个链表首部，长链表先走n步，然后两个指针同一时候走，直到两个指针所指向的值全然同样时停止。

代码：

/*
求链表公共结点
*/
#include<stdio.h>
#include<stdlib.h>
typedef struct _NODE_
{int data;struct _NODE_ *next;
}Node,*pNode;
int get_length(pNode list)
{if(list == NULL){return 0;}int len = 0;pNode pTemp = list;while(pTemp != NULL){len++;pTemp = pTemp->next;}return len;
}
void create(pNode *list,int n)
{if(n <= 0){return;}*list = (pNode)malloc(sizeof(Node));if(!list){exit(-1);}int data;scanf("%d",&data);(*list)->data = data;(*list)->next = NULL;pNode pTemp = *list;for(int i = 0; i < n-1; i++){scanf("%d",&data);pNode pNew = (pNode)malloc(sizeof(Node));if(!pNew){exit(-1);}pNew->data = data;pNew->next = NULL;pTemp->next = pNew;pTemp = pNew;}
}
//求单链表公共结点
pNode FindCommonNode(pNode list1,pNode list2)
{if(list1 == NULL || list2 == NULL){return NULL;}int len1 = get_length(list1);int len2 = get_length(list2);int dif = len1 - len2;pNode pLong = list1;pNode pShort = list2;if(dif < 0){pLong = list2;pShort = list1;dif = len2-len1;}while(dif > 0){pLong = pLong->next;dif--;}while(pLong != NULL && pShort != NULL && pLong != pShort){pLong = pLong->next;pShort = pShort->next;}return pLong;
}
void Destroy(pNode *list)
{if(*list == NULL){return;}pNode p = *list,q;while(p != NULL){q = p->next;free(p);p = q;}
}

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