剑指offer25--合并两个排序的链表——java/C++实现

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。

 

思路：

• 判断空
• 头节点需要判断，
• 然后让头节点一个一个吞噬链表。

比较麻烦

/*
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==NULL||pHead2==NULL)
            return pHead1==NULL?pHead2:pHead1;
        ListNode* tmp1=pHead1;
        ListNode* tmp2=pHead2;
        ListNode* front(0);
        ListNode* refront(0);
        if(tmp1->val>tmp2->val){
                front=tmp2;
                refront=front;
                tmp2=tmp2->next;
        }else{
                front=tmp1;
                refront=front;
                tmp1=tmp1->next;
        }
        while(tmp1!=NULL&&tmp2!=NULL)
        {
            if(tmp1->val>tmp2->val){
                ListNode* tmp=tmp2;
                tmp2=tmp2->next;
                front->next=tmp;
                front=front->next;
            }else{
                ListNode* tmp=tmp1;
                tmp1=tmp1->next;
                front->next=tmp;
                front=front->next;
            }
        }
        while(tmp1!=NULL){
            front->next=tmp1;
            tmp1=tmp1->next;
        }
        while(tmp2!=NULL){
            front->next=tmp2;
            tmp2=tmp2->next;
        }

        return refront;
    }
};

剑指offer解法：

递归的解法比较好用

/*
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==NULL)
            return pHead2;
        else if(pHead2==NULL)
            return pHead1;
        ListNode* tmp(0);
        if(pHead1->val>pHead2->val)
        {
            tmp=pHead2;
            tmp->next=Merge(pHead1,pHead2->next);
        }else{
            tmp=pHead1;
            tmp->next=Merge(pHead1->next,pHead2);
        }
        return tmp;
    }
};

JAVA解法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        ListNode head = null;
        ListNode pre = null;
        if(l1.val <= l2.val){
            head = l1;
            pre = l1;
            l1 = l1.next;
        }else{
            head = l2;
            pre = l2;
            l2 = l2.next;
        }
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                pre.next = l1;
                l1 = l1.next;                
            }else{
                pre.next = l2;
                l2 = l2.next;
            }
            pre = pre.next;
        }
        while(l1 != null){
            pre.next = l1;
            pre = pre.next;
            l1 = l1.next;
        }
        while(l2 != null){
            pre.next = l2;
            pre = pre.next;
            l2 = l2.next;
        }
        return head;
    }
}