《剑指offer》16：合并两个排序的列表

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++实现：

/*
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)
            return pHead2;
        if(!pHead2)
            return pHead1;
        
        ListNode *c1 = NULL, *c2 = NULL, *t = NULL, *p1 = NULL, *p2 = NULL, *r = NULL;
        c1 = pHead1;
        c2 = pHead2;
        
        if(c1 ->val >= c2 -> val)
        {
            t = c2;
            p2 = c2 -> next;
            p1 = c1;
        }
        else
        {
            t = c1;
            p1 = c1 -> next;
            p2 = c2;
        }
        
        r = t;
        while(p1 != NULL && p2 != NULL)
        {
            if(p1 ->val >= p2 -> val)
            {
                t -> next = p2;
                t = p2;
                p2 = p2 -> next;
            }
            else
            {
                t -> next = p1;
                t = p1;
                p1 = p1 -> next;
            }
        }
        
        if(p2 == NULL)
        {
            t -> next = p1;
        }
        
        if(p1 == NULL)
        {
            t -> next = p2;
        }
        
        return r;
    }
};

python实现：

# -*- 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):
        # write code here
        t = ListNode(0)
        r = t
        if pHead1 == None:
            return pHead2
        if pHead2 == None:
            return pHead1
        
        if pHead1.val <= pHead2.val:
            t = pHead1
            pHead1 = pHead1.next
        else:
            t = pHead2
            pHead2 = pHead2.next
        
        r = t
        
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                t.next = pHead1;
                t = pHead1;
                pHead1 = pHead1.next
            else:
                t.next = pHead2;
                t = pHead2;
                pHead2 = pHead2.next
                
        if pHead1 == None:
            t.next = pHead2
        if pHead2 == None:
            t.next = pHead1
        
        return r