《剑指offer》第二十五题:合并两个排序的链表
// 面试题25:合并两个排序的链表 // 题目:输入两个递增排序的链表,合并这两个链表并使新链表中的结点仍然是按 // 照递增排序的。例如输入图3.11中的链表1和链表2,则合并之后的升序链表如链 // 表3所示。 #include <cstdio> #include "List.h" ListNode* Merge(ListNode* pHead1, ListNode* pHead2) { if (pHead1 == nullptr) return pHead2; else if (pHead2 == nullptr) return pHead1; ListNode* pNewHead = nullptr; if (pHead1->m_nValue < pHead2->m_nValue) { pNewHead = pHead1; pNewHead->m_pNext = Merge(pHead1->m_pNext, pHead2); } else { pNewHead = pHead2; pNewHead->m_pNext = Merge(pHead1, pHead2->m_pNext); } return pNewHead; }
// ====================测试代码==================== ListNode* Test(const char* testName, ListNode* pHead1, ListNode* pHead2) { if (testName != nullptr) printf("%s begins:\n", testName); printf("The first list is:\n"); PrintList(pHead1); printf("The second list is:\n"); PrintList(pHead2); printf("The merged list is:\n"); ListNode* pMergedHead = Merge(pHead1, pHead2); PrintList(pMergedHead); printf("\n\n"); return pMergedHead; } // list1: 1->3->5 // list2: 2->4->6 void Test1() { ListNode* pNode1 = CreateListNode(1); ListNode* pNode3 = CreateListNode(3); ListNode* pNode5 = CreateListNode(5); ConnectListNodes(pNode1, pNode3); ConnectListNodes(pNode3, pNode5); ListNode* pNode2 = CreateListNode(2); ListNode* pNode4 = CreateListNode(4); ListNode* pNode6 = CreateListNode(6); ConnectListNodes(pNode2, pNode4); ConnectListNodes(pNode4, pNode6); ListNode* pMergedHead = Test("Test1", pNode1, pNode2); DestroyList(pMergedHead); } // 两个链表中有重复的数字 // list1: 1->3->5 // list2: 1->3->5 void Test2() { ListNode* pNode1 = CreateListNode(1); ListNode* pNode3 = CreateListNode(3); ListNode* pNode5 = CreateListNode(5); ConnectListNodes(pNode1, pNode3); ConnectListNodes(pNode3, pNode5); ListNode* pNode2 = CreateListNode(1); ListNode* pNode4 = CreateListNode(3); ListNode* pNode6 = CreateListNode(5); ConnectListNodes(pNode2, pNode4); ConnectListNodes(pNode4, pNode6); ListNode* pMergedHead = Test("Test2", pNode1, pNode2); DestroyList(pMergedHead); } // 两个链表都只有一个数字 // list1: 1 // list2: 2 void Test3() { ListNode* pNode1 = CreateListNode(1); ListNode* pNode2 = CreateListNode(2); ListNode* pMergedHead = Test("Test3", pNode1, pNode2); DestroyList(pMergedHead); } // 一个链表为空链表 // list1: 1->3->5 // list2: 空链表 void Test4() { ListNode* pNode1 = CreateListNode(1); ListNode* pNode3 = CreateListNode(3); ListNode* pNode5 = CreateListNode(5); ConnectListNodes(pNode1, pNode3); ConnectListNodes(pNode3, pNode5); ListNode* pMergedHead = Test("Test4", pNode1, nullptr); DestroyList(pMergedHead); } // 两个链表都为空链表 // list1: 空链表 // list2: 空链表 void Test5() { ListNode* pMergedHead = Test("Test5", nullptr, nullptr); } int main(int argc, char* argv[]) { Test1(); Test2(); Test3(); Test4(); Test5(); return 0; }
分析:递归简洁之美。
/* struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { } };*/ class Solution { public: ListNode* Merge(ListNode* pHead1, ListNode* pHead2) { if (pHead1 == nullptr) return pHead2; else if (pHead2 == nullptr) return pHead1; ListNode* pMergeHead = nullptr; if (pHead1->val < pHead2->val) { pMergeHead = pHead1; pMergeHead->next = Merge(pHead1->next, pHead2); } else { pMergeHead = pHead2; pMergeHead->next = Merge(pHead1, pHead2->next); } return pMergeHead; } };