将两个有序链表融合成一个有序链表

已知两个有序链表，将两个有序链表融合成一个有序链表的实现方法：

#include <iostream>using namespace std;typedef struct _listNode{int data;struct _listNode *next;}listNode;void printList(listNode *plist){if(plist == NULL)return;listNode *tlist = plist;while(tlist != NULL){cout << tlist->data << "\t";tlist = tlist->next;}cout << endl;}listNode *listNodeMerge(listNode *p1ist1, listNode *p1ist2){listNode *p1 = p1ist1;listNode *p2 = p1ist2;listNode *pcurr = NULL;listNode *pHead = NULL;listNode *pt = NULL;//先判断链表是否为空if(p1 == NULL && p2 != NULL)return p2;else if(p1 != NULL && p2 == NULL)return p1;else if(p1 == NULL && p2 == NULL)return NULL;//首先选择一个头节点if(p1->data < p2->data){pHead = p1;p1 = p1->next;}else{pHead = p2;p2 = p2->next;}pcurr = pHead; //辅助指针while(p1 != NULL && p2 != NULL){if(p1->data < p2->data){pt = p1;p1 = p1->next;}else{pt = p2;p2 = p2->next;}pcurr->next = pt;pcurr = pcurr->next;}if(p1 != NULL) //其中有一个链表已经指向NULL，而另一个还没有指向NULLpcurr->next = p1;elsepcurr->next = p2;return pHead;}int main(){//为方便先定义两个有序链表，也可用new(malloc)的方式生成链表listNode n1,n3,n5;listNode n2,n4,n6,n7;n1.data = 1, n1.next = &n3;n3.data = 2, n3.next = &n5;n5.data = 3, n5.next = NULL;n2.data = 0, n2.next = &n4;n4.data = 4, n4.next = &n6;n6.data = 6, n6.next = &n7;n7.data = 9, n7.next = NULL;listNode *p1 = &n1;listNode *p2 = &n2;//打印该链表cout << "list1:\n";printList(p1);cout << "list2:\n";printList(p2);//打印融合后的链表cout << "After merge list:\n";listNode *pnew = listNodeMerge(p1, p2);printList(pnew);return 0;}