使用 C++ 合并两个已排序的链表。
问题陈述
已知两个按序排列的单链表。编写一个函数以合并给定的两个已排序的链表
List1: 10->15->17->20 List2: 5->9->13->19 Result: 5->9->10->13->15->17->19->20
算法
1. Traverse both lists 1.1. If list1->data < list2->data 1.1.1 Add list1->data to new list and increment list1 pointer 1.2 If list2->data < list1->data 1.2.1 Add list2->data to new list and increment list2 pointer 2. Repeat procedure until both lists are exhausted 3. Return resultant list
示例
#include <iostream> #include <new> #define SIZE(arr) (sizeof(arr) / sizeof(arr[0])) using namespace std; struct node { int data; struct node *next; }; node *createList(int *arr, int n){ node *head, *p; p = head = new node; head->data = arr[0]; head->next = NULL; for (int i = 1; i < n; ++i) { p->next = new node; p = p->next; p->data = arr[i]; p->next = NULL; } return head; } void displayList(node *head){ while (head != NULL) { cout << head->data << " "; head = head->next; } cout << endl; } node *mergeSortedLists(node *head1, node *head2){ node *result = NULL; if (head1 == NULL) { return head2; } if (head2 == NULL) { return head1; } if (head1->data < head2->data) { result = head1; result->next = mergeSortedLists(head1->next, head2); } else { result = head2; result->next = mergeSortedLists(head1, head2->next); } return result; } int main(){ int arr1[] = {10, 15, 17, 20}; int arr2[] = {5, 9, 13, 19}; node *head1, *head2, *result = NULL; head1 = createList(arr1, SIZE(arr1)); head2 = createList(arr2, SIZE(arr1)); cout << "First sorted list: " << endl; displayList(head1); cout << "Second sorted list: " << endl; displayList(head2); result = mergeSortedLists(head1, head2); cout << "Final sorted list: " << endl; displayList(result); return 0; }
输出
当你编译并执行上述程序时,它将生成以下输出 -
First sorted list: 10 15 17 20 Second sorted list: 5 9 13 19 Final sorted list: 5 9 10 13 15 17 19 20
广告