剑指offer66题--Java实现，c++实现和python实现 16.合并两个排序的链表

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

C++

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode* node=NULL;
        if(pHead1==NULL){return node=pHead2;}
        if(pHead2==NULL){return node=pHead1;}
        if(pHead1->val>pHead2->val){
            node=pHead2;
            node->next=Merge(pHead1,pHead2->next);
        }else
            {
            node=pHead1;
            node->next=Merge(pHead1->next,pHead2);
        }
        return node;
    }
};

Java

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        ListNode newHead = new ListNode(-1);
        ListNode current = newHead;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                current.next = list1;
                list1 = list1.next;
            } else {
                current.next = list2;
                list2 = list2.next;
            }
            current = current.next;
        }
        if (list1 != null) current.next = list1;
        if (list2 != null) current.next = list2;
        return newHead.next;
    }
}

python

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        if not pHead1:
            return pHead2
        if not pHead2:
            return  pHead1
        if pHead1.val <= pHead2.val:
            pHead1.next = self.Merge(pHead1.next, pHead2)
            return pHead1
        else:
            pHead2.next = self.Merge(pHead1, pHead2.next)
            return pHead2