合并两个有序链表

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方法1:递归求解

首先设两个单链表的头节点分别为head1、head2。如果head1为空，则直接返回head2，若head2为空，则直接返回head1。若head1与head2都不为空，则比较head1-.val 与head2->val的大小，来决定head1和head2谁为合并后的单链表的头节点。分两种情况：

1.若head1->val <= head2->val，则头节点head = head1,接着递归去求分别以head1->next 和 head2 为头节点的两个单链表的合并链表的子问题。

2.若head1->val > head2->val，则头节点head = head2,接着递归去求分别以head1 和 head2->next 为头节点的两个单链表的合并链表的子问题。

方法2：非递归求解

首先分别用指针head1，head2来遍历两个链表，如果当前head1->val <= head2->val，则将head1指向的节点归入到合并后的链表中；否则将head2指向的节点归入到合并后的链表中。当有一个链表遍历结束，则把尚未遍历完毕的链表直接链接到合并的链表尾部即可。

以下是上述两种方法的代码实现：

#include <iostream>#include <vector>#include <map>using namespace std;struct listNode{int val;struct listNode *next;};// 方法1：递归法,返回一个指向排序后的链表的头指针listNode * mergeRecursive(listNode *head1, listNode *head2){if (head1 == NULL){return head2;}if (head2 == NULL){return head1;}listNode *head = NULL;if (head1->val <= head2->val){head = head1;head->next = mergeRecursive(head1->next, head2);}else{head = head2;head->next = mergeRecursive(head1, head2->next);}return head;}// 方法2：非递归法求解listNode *mergeListNode(listNode *head1, listNode *head2){listNode *cur = new listNode;cur->val = 999;cur->next = NULL;listNode *rst = cur;while (head1 != NULL && head2 != NULL){if (head1->val <= head2->val){cur->next = head1;cur = head1;head1 = head1->next;}else{cur->next = head2;cur = head2;head2 = head2->next;}}if (head1 != NULL){cur->next = head1;}if (head2 != NULL){cur->next = head2;}return rst->next;}// 打印单链表void printList(listNode *head){listNode *cur = head;while(cur != NULL){cout << cur->val << " ";cur = cur->next;}cout << endl;}int main(){// 先建立一个有6个节点的单链表1：0-1-3-5-7-9listNode *head1 = new listNode;head1->val = 0;int n = 1;listNode *cur1 = head1;while(n <= 9){cur1->next = new listNode;cur1 = cur1->next;cur1->val = n;n += 2;}cur1->next = NULL;cur1 = head1;n = 0;// 建立一个有6个节点的单链表2: -1-0-2-4-8-10listNode *head2 = new listNode;head2->val = -1;n = 0;listNode *cur2 = head2;while (n <=10){cur2->next = new listNode;cur2 = cur2->next;cur2->val = n;n += 2;}cur2->next = NULL;cur2 = head2;// 打印两单链表1,2printList(cur1);printList(cur2);listNode *result = mergeRecursive(cur1, cur2);cur1 = head1;cur2 = head2;cout << "利用递归法合并后的单链表为：" ;printList(result);// 先建立一个有6个节点的单链表3：0-1-3-5-7-9listNode *head3 = new listNode;head3->val = 0;int nn = 1;listNode *cur3 = head3;while(nn <= 9){cur3->next = new listNode;cur3 = cur3->next;cur3->val = nn;nn += 2;}cur3->next = NULL;cur3 = head3;// 建立一个有6个节点的单链表4: -1-0-2-4-8-10listNode *head4 = new listNode;head4->val = -1;nn = 0;listNode *cur4 = head4;while (nn <=10){cur4->next = new listNode;cur4 = cur4->next;cur4->val = nn;nn += 2;}cur4->next = NULL;cur4 = head4;// 打印两单链表3,4printList(cur3);printList(cur4);result = mergeListNode(cur3, cur4);cout << "利用非递归法合并后的单链表为：" ;printList(result);return 0;}

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