33.两个链表的第一个公共结点

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两个链表的第一个公共结点

参与人数：2623时间限制：1秒空间限制：32768K
本题知识点：链表
算法知识视频讲解

题目描述

输入两个链表，找出它们的第一个公共结点。

最喜欢做链表的题了，舒爽~

// 31.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}};class Solution {public:ListNode* FindFirstCommonNode(ListNode *pHead1, ListNode *pHead2) {if (pHead1 == NULL || pHead2 == NULL) return NULL;int length1 = 0;ListNode* pNode1 = pHead1;while (pNode1 != NULL) {pNode1 = pNode1->next;length1++;}int length2 = 0;ListNode* pNode2 = pHead2;while (pNode2 != NULL) {pNode2 = pNode2->next;length2++;}int cur = length2;pNode1 = pHead1;pNode2 = pHead2;if (length1 > length2) {int n = length1 - length2;for (int i = 0; i < n; i++) {pNode1 = pNode1->next;}cur = length2;}else if (length2 > length1) {int n = length2 - length1;for (int i = 0; i < n; i++) {pNode2 = pNode2->next;}cur = length1;}ListNode* retVal = NULL;for (int i = 0; i < cur; i++) {if (pNode1 == pNode2) {retVal = pNode1;break;}pNode1 = pNode1->next;pNode2 = pNode2->next;}return retVal;}};int _tmain(int argc, _TCHAR* argv[]){ListNode p1(1);ListNode p2(2);ListNode p3(3);p1.next = &p2;p2.next = &p3;ListNode p4(4);ListNode p5(5);p4.next = &p5;ListNode p6(6);ListNode p7(7);p6.next = &p7;p3.next = &p6;p5.next = &p6;Solution s;ListNode* result = s.FindFirstCommonNode(&p1, &p4);return 0;}

</pre><pre code_snippet_id="1678311" snippet_file_name="blog_20160510_1_4105559" name="code" class="cpp">第二次做：

<pre name="code" class="cpp">/*struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}};*/class Solution {public:    ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) {        if ( pHead1 == NULL || pHead2 == NULL ) return NULL ;                int length1 = 0 ;        int length2 = 0 ;                ListNode* pNode1 = pHead1 ;        while ( pNode1 != NULL ) {            ++ length1 ;            pNode1 = pNode1->next ;        }                ListNode* pNode2 = pHead2 ;        while ( pNode2 != NULL ) {            ++ length2 ;            pNode2 = pNode2->next ;        }                pNode1 = pHead1 ;        pNode2 = pHead2 ;        if ( length1 > length2 ) {            int gap = length1 - length2 ;            for ( int i = 0; i < gap; ++ i ) {                pNode1 = pNode1->next ;            }        } else if ( length1 < length2 ) {            int gap = length2 - length1 ;            for ( int i = 0; i < gap; ++ i ) {                pNode2 = pNode2->next ;            }        }                ListNode* retNode = NULL ;        while ( pNode1 != NULL && pNode2 != NULL ) {            if ( pNode1 == pNode2 ) {                retNode = pNode1 ;                break ;            }            pNode1 = pNode1->next ;            pNode2 = pNode2->next ;        }        return retNode ;    }};

第三次做：

/*struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}};*/class Solution {public:    ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) {        if ( pHead1 == NULL || pHead2 == NULL ) return NULL ;                int length1 = 0 ;        int length2 = 0 ;        ListNode* pNode1 = pHead1 ;        ListNode* pNode2 = pHead2 ;        while ( pNode1 != NULL ) {            ++ length1 ;            pNode1 = pNode1->next ;        }        while ( pNode2 != NULL ) {            ++ length2 ;            pNode2 = pNode2->next ;        }                pNode1 = pHead1 ;        pNode2 = pHead2 ;        if ( length1 > length2 ) {            int step = length1 - length2 ;            for ( int i = 0; i < step; ++ i ) {                pNode1 = pNode1->next ;            }        } else {            int step = length2 - length1 ;            for ( int i = 0; i < step; ++ i ) {                pNode2 = pNode2->next ;            }        }                while ( pNode1 != NULL && pNode2 != NULL ) {            if ( pNode1 == pNode2 ) return pNode1 ;            pNode1 = pNode1->next ;            pNode2 = pNode2->next ;        }                return NULL ;    }};

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