C++刷剑指offer(面试题25. 合并两个排序的链表）【链表】
本帖最后由 糖逗 于 2020-5-8 17:31 编辑题目描述：
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节点仍然是递增排序的。
示例1：
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4
限制：
0 <= 链表长度 <= 1000
#include <iostream>
using namespace std;
struct ListNode{
int value;
ListNode* next;
ListNode(int x): value(x), next(NULL){
}
};
void printList(ListNode* input){
ListNode* temp = input;
while(temp -> next){
temp = temp -> next;
cout << temp -> value << " ";
}
cout << endl;
cout << "-------------" << endl;
}
ListNode* solution(ListNode* input1, ListNode* input2){
ListNode* temp1 = input1 -> next;
ListNode* temp2 = input2 -> next;
ListNode* temp = new ListNode(0);
ListNode* res = temp;
while(temp1 != NULL || temp2 != NULL){
if(temp1 == NULL){
temp -> next = temp2;
break;
}
if(temp2 == NULL){
temp -> next = temp1;
break;
}
if(temp1 -> value <=temp2 -> value){
temp -> next = temp1;
temp = temp -> next;
temp1 = temp1 -> next;
continue;
}
if(temp1 -> value > temp2 -> value){
temp -> next = temp2;
temp = temp -> next;
temp2 = temp2 -> next;
continue;
}
}
return res;
}
int main(void){
ListNode* input1 = new ListNode(0);
ListNode* temp1 = input1;
cout << "please send numbers for the first singleList:" << endl;
int number1;
while(cin >> number1){
ListNode* node = new ListNode(number1);
temp1 -> next = node;
temp1 = node;
}
printList(input1);
cin.clear();
ListNode* input2 = new ListNode(0);
ListNode* temp2 = input2;
cout << "please send numbers for the second singleList:" << endl;
int number2;
while(cin >> number2){
ListNode* node = new ListNode(number2);
temp2 -> next = node;
temp2 = node;
}
printList(input2);
cin.clear();
cout << "start!" << endl;
ListNode* res = solution(input1, input2);
cout << "finish!" << endl;
printList(res);
return 0;
}
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