Programming Algorithm-code of the first public node of two linked lists (c)

Code of the first public node of the two linked lists (c)

Address: http://blog.csdn.net/caroline_wendy

Question: enter two linked lists to find their first public node.

Calculate the length of the linked list, move the pointer of a longer-chain table to the same distance to the same node, move the pointer of the two linked lists at the same time, and find the same element.

Time Complexity: O (N)

Code:

/* * main.cpp * *  Created on: 2014.6.12 *      Author: Spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <stdlib.h>#include <string.h>struct ListNode {int m_nKey;ListNode* m_pNext;};size_t GetListLength (ListNode* pHead) {size_t nLength = 0;ListNode* pNode = pHead;while (pNode != NULL) {++nLength;pNode = pNode->m_pNext;}return nLength;}ListNode* FindFirstCommonNode(ListNode* pHead1, ListNode* pHead2) {size_t nLength1 = GetListLength(pHead1);size_t nLength2 = GetListLength(pHead2);int nLengthDif = nLength1 - nLength2;ListNode* pListHeadLong = pHead1;ListNode* pListHeadShort = pHead2;if (nLength2 > nLength1) {pListHeadLong = pHead2;pListHeadShort = pHead1;nLengthDif = nLength2 - nLength1;}for (int i=0; i<nLengthDif; ++i)pListHeadLong = pListHeadLong->m_pNext;while ((pListHeadLong != NULL) && (pListHeadShort != NULL)&& (pListHeadLong != pListHeadShort)) {pListHeadLong = pListHeadLong->m_pNext;pListHeadShort = pListHeadShort->m_pNext;}ListNode* pFirstCommonNode = pListHeadLong;return pFirstCommonNode;}int main(void){ListNode* pHead1 = new ListNode();ListNode* pHead1Node1 = new ListNode();ListNode* pHead1Node2 = new ListNode();ListNode* pHead1Node3 = new ListNode();ListNode* pHead1Node4 = new ListNode();pHead1->m_nKey = 1;pHead1Node1->m_nKey = 2;pHead1Node2->m_nKey = 3;pHead1Node3->m_nKey = 6;pHead1Node4->m_nKey = 7;pHead1->m_pNext = pHead1Node1;pHead1Node1->m_pNext = pHead1Node2;pHead1Node2->m_pNext = pHead1Node3;pHead1Node3->m_pNext = pHead1Node4;pHead1Node4->m_pNext = NULL;ListNode* pHead2 = new ListNode();ListNode* pHead2Node1 = new ListNode();pHead2->m_nKey = 4;pHead2Node1->m_nKey = 5;pHead2->m_pNext = pHead2Node1;pHead2Node1->m_pNext = pHead1Node3;ListNode* result = FindFirstCommonNode(pHead1, pHead2);printf("result = %d\n", result->m_nKey);    return 0;}

Output:

result = 6