剑指offer第十七题：合并两个排序的链表
1 //============================================================================ 2 // Name : JZ-C-17.cpp 3 // Author : Laughing_Lz 4 // Version : 5 // Copyright : All Right Reserved 6 // Description :合并两个排序的链表 7 //============================================================================ 8 9 #include <iostream> 10 #include <stdio.h> 11 #include "List.h" 12 using namespace std; 13 ListNode* Merge(ListNode* pHead, ListNode* qHead) { 14 if (pHead == NULL) { //考虑某链表为空的情况 15 return qHead; 16 } else if (qHead == NULL) { 17 return pHead; 18 } 19 ListNode* NewHead = NULL; 20 if (pHead->m_nValue < qHead->m_nValue) { 21 NewHead = pHead; 22 NewHead->m_pNext = Merge(pHead->m_pNext, qHead); //递归 23 } else { //包含了相等的情况 24 NewHead = qHead; 25 NewHead->m_pNext = Merge(pHead, qHead->m_pNext); 26 } 27 return NewHead; 28 } 29 30 // ====================测试代码==================== 31 ListNode* Test(char* testName, ListNode* pHead1, ListNode* pHead2) { 32 if (testName != NULL) 33 printf("%s begins: ", testName); 34 35 printf("The first list is: "); 36 PrintList(pHead1); 37 38 printf("The second list is: "); 39 PrintList(pHead2); 40 41 printf("The merged list is: "); 42 ListNode* pMergedHead = Merge(pHead1, pHead2); 43 PrintList(pMergedHead); 44 45 printf(" "); 46 47 return pMergedHead; 48 } 49 50 // list1: 1->3->5 51 // list2: 2->4->6 52 void Test1() { 53 ListNode* pNode1 = CreateListNode(1); 54 ListNode* pNode3 = CreateListNode(3); 55 ListNode* pNode5 = CreateListNode(5); 56 57 ConnectListNodes(pNode1, pNode3); 58 ConnectListNodes(pNode3, pNode5); 59 60 ListNode* pNode2 = CreateListNode(2); 61 ListNode* pNode4 = CreateListNode(4); 62 ListNode* pNode6 = CreateListNode(6); 63 64 ConnectListNodes(pNode2, pNode4); 65 ConnectListNodes(pNode4, pNode6); 66 67 ListNode* pMergedHead = Test("Test1", pNode1, pNode2); 68 69 DestroyList(pMergedHead); 70 } 71 72 // 两个链表中有重复的数字 73 // list1: 1->3->5 74 // list2: 1->3->5 75 void Test2() { 76 ListNode* pNode1 = CreateListNode(1); 77 ListNode* pNode3 = CreateListNode(3); 78 ListNode* pNode5 = CreateListNode(5); 79 80 ConnectListNodes(pNode1, pNode3); 81 ConnectListNodes(pNode3, pNode5); 82 83 ListNode* pNode2 = CreateListNode(1); 84 ListNode* pNode4 = CreateListNode(3); 85 ListNode* pNode6 = CreateListNode(5); 86 87 ConnectListNodes(pNode2, pNode4); 88 ConnectListNodes(pNode4, pNode6); 89 90 ListNode* pMergedHead = Test("Test2", pNode1, pNode2); 91 92 DestroyList(pMergedHead); 93 } 94 95 // 两个链表都只有一个数字 96 // list1: 1 97 // list2: 2 98 void Test3() { 99 ListNode* pNode1 = CreateListNode(1); 100 ListNode* pNode2 = CreateListNode(2); 101 102 ListNode* pMergedHead = Test("Test3", pNode1, pNode2); 103 104 DestroyList(pMergedHead); 105 } 106 107 // 一个链表为空链表 108 // list1: 1->3->5 109 // list2: 空链表 110 void Test4() { 111 ListNode* pNode1 = CreateListNode(1); 112 ListNode* pNode3 = CreateListNode(3); 113 ListNode* pNode5 = CreateListNode(5); 114 115 ConnectListNodes(pNode1, pNode3); 116 ConnectListNodes(pNode3, pNode5); 117 118 ListNode* pMergedHead = Test("Test4", pNode1, NULL); 119 120 DestroyList(pMergedHead); 121 } 122 123 // 两个链表都为空链表 124 // list1: 空链表 125 // list2: 空链表 126 void Test5() { 127 ListNode* pMergedHead = Test("Test5", NULL, NULL); 128 } 129 130 int main(int argc, char** argv) { 131 Test1(); 132 Test2(); 133 Test3(); 134 Test4(); 135 Test5(); 136 137 return 0; 138 }