合併兩個有序鏈表，保持鏈表順序。

（編程題）合併兩個有序鏈表，保持鏈表順序。

例如:

輸入:

鏈表1: 1->3->5->7

鏈表2: 2->4->6->8

輸出:

鏈表交集: 1->2->3->4->5->6->7->8

一般做法：

/*
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;
        if(pHead2 == nullptr)
            return pHead1;
        //ListNode resHead(0);
        ListNode* dummy = new ListNode(-1);//創建新的頭結點
        ListNode* p = dummy;//p指向dummy，p指針向下移動，一直更新 
        while(pHead1 && pHead2){
            if(pHead1->val < pHead2->val){
                p->next = pHead1;//當前較小值爲p1，連接到p指針後面
                pHead1 = pHead1->next;//p1指針後移一位
            }
            else{
                p->next = pHead2;
                pHead2 = pHead2->next;
            }
            p = p->next;//p指針後移
        }
        if(pHead1)//如果最後剩下的是p1鏈表，則把p1連接到p上
            p->next = pHead1;
        if(pHead2)//如果最後剩下的是p2鏈表，則把p2連接到p上
            p->next = pHead2;
        return dummy -> next;
    }
};

測試程序：

不一般做法：

#include <iostream>
#include <vector> 
#include <list>
//#include <bits/stdc++.h>
using namespace std;
 
struct ListNode{
	int val;
	struct ListNode *next;
	ListNode(int x):val(x),next(NULL){
	}

};

ListNode* Merge(ListNode* l1, ListNode* l2){
	if(l1 == nullptr)
		return l2;
	if(l2 == nullptr)
		return l1;
		
	ListNode* dummy = new ListNode(-1);
	ListNode* cur = dummy;
	while(l1 && l2){
		if(l1->val < l2->val){
			cur->next = l1;
			l1 = l1->next;
		}
		else{
			cur->next = l2;
			l2 = l2->next;
		}
		cur = cur->next;
	}
	if(l1)
		cur->next = l1;
	if(l2)
		cur->next = l2;
		
	return dummy->next;
}

ListNode* createList(vector<int>& num){
	if(num.size()== 0) return nullptr;
	ListNode* head = new ListNode(num[0]);
	ListNode* cur = head;
	for(int i = 1; i <num.size(); i++){
		cur->next = new ListNode(num[i]);
		cur = cur->next;
	}
	return head;
}

int main(int argc, char** argv) {
	vector<int> num1,num2;
	int n1, n2;
	while(cin>>n1){
		num1.push_back(n1);
		if(cin.get() == '\n')
			break;
	}
	while(cin>>n2){
		num2.push_back(n2);
		if(cin.get() == '\n')
			break;
	}
	
	ListNode* head1 = createList(num1);
	ListNode* head2 = createList(num2);
	ListNode* res = Merge(head1, head2);
	ListNode* p = res;
	while(p){
		cout<< p->val<<" ";
		p = p->next;
	}
	
	return 0;
}

 

【數組】

鏈接：https://www.nowcoder.com/questionTerminal/27c833289e5f4f5e9ba3718ce9136759
來源：牛客網

#include <bits/stdc++.h>
using namespace std;
int main()
{
    vector<int> nums1, nums2,nums;
    int num1, num2;
    while(cin >> num1)
    {
        nums1.push_back(num1);
        nums.push_back(num1);
        if(cin.get() == '\n')
            break;
    }
    while(cin >> num2)
    {
        nums2.push_back(num2);
        nums.push_back(num2);
        if(cin.get() == '\n')
            break;
    }
    sort(nums.begin(),nums.end());
    for(int i=0;i<nums.size();i++)
        cout<<nums[i]<<" ";
    return 0;
}