两个链表的第一个公共结点

来源：互联网发布：小票打印机软件编辑：程序博客网时间：2024/05/30 23:48
问题:输入两个链表，找出它们的第一个公共结点。链表的结点定义如下:
struct ListNode
{
int m_nKey;
ListNode * m_pNext;
}
解决问题的思路:
(1)蛮力法：时间复杂度O(m*n)
(2)运用栈的反向思考
ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2){    // 得到两个链表的长度    unsigned int nLength1 = GetListLength(pHead1);    unsigned int nLength2 = GetListLength(pHead2);    int nLengthDif = nLength1 - nLength2;     ListNode* pListHeadLong = pHead1;    ListNode* pListHeadShort = pHead2;    if(nLength2 > nLength1)    {        pListHeadLong = pHead2;        pListHeadShort = pHead1;        nLengthDif = nLength2 - nLength1;    }     // 先在长链表上走几步，再同时在两个链表上遍历    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;    }     // 得到第一个公共结点    ListNode* pFisrtCommonNode = pListHeadLong;     return pFisrtCommonNode;}unsigned int GetListLength(ListNode* pHead){    unsigned int nLength = 0;    ListNode* pNode = pHead;    while(pNode != NULL)    {        ++ nLength;        pNode = pNode->m_pNext;    }     return nLength;}// ====================测试代码====================void DestroyNode(ListNode* pNode);void Test(char* testName, ListNode* pHead1, ListNode* pHead2, ListNode* pExpected){    if(testName != NULL)        printf("%s begins: ", testName);    ListNode* pResult = FindFirstCommonNode(pHead1, pHead2);    if(pResult == pExpected)        printf("Passed.\n");    else        printf("Failed.\n");}// 第一个公共结点在链表中间// 1 - 2 - 3 \//            6 - 7//     4 - 5 /void Test1(){    ListNode* pNode1 = CreateListNode(1);    ListNode* pNode2 = CreateListNode(2);    ListNode* pNode3 = CreateListNode(3);    ListNode* pNode4 = CreateListNode(4);    ListNode* pNode5 = CreateListNode(5);    ListNode* pNode6 = CreateListNode(6);    ListNode* pNode7 = CreateListNode(7);    ConnectListNodes(pNode1, pNode2);    ConnectListNodes(pNode2, pNode3);    ConnectListNodes(pNode3, pNode6);    ConnectListNodes(pNode4, pNode5);    ConnectListNodes(pNode5, pNode6);    ConnectListNodes(pNode6, pNode7);    Test("Test1", pNode1, pNode4, pNode6);    DestroyNode(pNode1);    DestroyNode(pNode2);    DestroyNode(pNode3);    DestroyNode(pNode4);    DestroyNode(pNode5);    DestroyNode(pNode6);    DestroyNode(pNode7);}// 没有公共结点// 1 - 2 - 3 - 4//            // 5 - 6 - 7void Test2(){    ListNode* pNode1 = CreateListNode(1);    ListNode* pNode2 = CreateListNode(2);    ListNode* pNode3 = CreateListNode(3);    ListNode* pNode4 = CreateListNode(4);    ListNode* pNode5 = CreateListNode(5);    ListNode* pNode6 = CreateListNode(6);    ListNode* pNode7 = CreateListNode(7);    ConnectListNodes(pNode1, pNode2);    ConnectListNodes(pNode2, pNode3);    ConnectListNodes(pNode3, pNode4);    ConnectListNodes(pNode5, pNode6);    ConnectListNodes(pNode6, pNode7);    Test("Test2", pNode1, pNode5, NULL);    DestroyList(pNode1);    DestroyList(pNode5);}// 公共结点是最后一个结点// 1 - 2 - 3 - 4 \//                7//         5 - 6 /void Test3(){    ListNode* pNode1 = CreateListNode(1);    ListNode* pNode2 = CreateListNode(2);    ListNode* pNode3 = CreateListNode(3);    ListNode* pNode4 = CreateListNode(4);    ListNode* pNode5 = CreateListNode(5);    ListNode* pNode6 = CreateListNode(6);    ListNode* pNode7 = CreateListNode(7);    ConnectListNodes(pNode1, pNode2);    ConnectListNodes(pNode2, pNode3);    ConnectListNodes(pNode3, pNode4);    ConnectListNodes(pNode4, pNode7);    ConnectListNodes(pNode5, pNode6);    ConnectListNodes(pNode6, pNode7);    Test("Test3", pNode1, pNode5, pNode7);    DestroyNode(pNode1);    DestroyNode(pNode2);    DestroyNode(pNode3);    DestroyNode(pNode4);    DestroyNode(pNode5);    DestroyNode(pNode6);    DestroyNode(pNode7);}// 公共结点是第一个结点// 1 - 2 - 3 - 4 - 5// 两个链表完全重合   void Test4(){    ListNode* pNode1 = CreateListNode(1);    ListNode* pNode2 = CreateListNode(2);    ListNode* pNode3 = CreateListNode(3);    ListNode* pNode4 = CreateListNode(4);    ListNode* pNode5 = CreateListNode(5);    ConnectListNodes(pNode1, pNode2);    ConnectListNodes(pNode2, pNode3);    ConnectListNodes(pNode3, pNode4);    ConnectListNodes(pNode4, pNode5);    Test("Test4", pNode1, pNode1, pNode1);    DestroyList(pNode1);}// 输入的两个链表有一个空链表void Test5(){    ListNode* pNode1 = CreateListNode(1);    ListNode* pNode2 = CreateListNode(2);    ListNode* pNode3 = CreateListNode(3);    ListNode* pNode4 = CreateListNode(4);    ListNode* pNode5 = CreateListNode(5);    ConnectListNodes(pNode1, pNode2);    ConnectListNodes(pNode2, pNode3);    ConnectListNodes(pNode3, pNode4);    ConnectListNodes(pNode4, pNode5);    Test("Test5", NULL, pNode1, NULL);    DestroyList(pNode1);}// 输入的两个链表有一个空链表void Test6(){    Test("Test6", NULL, NULL, NULL);}void DestroyNode(ListNode* pNode){    delete pNode;    pNode = NULL;}int _tmain(int argc, _TCHAR* argv[]){    Test1();    Test2();    Test3();    Test4();    Test5();    Test6();    return 0;}
总结:降低时间复杂度的两种策略:
(1)设计更加高效的算法
(2)用空间换取时间
总之，具体问题具体分析.
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