一.题目

Merge Two Sorted Lists

Total Accepted: 63974 Total Submissions: 196044My
Submissions

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

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二.解题技巧

    这道题就是将两个已排序的列表的元素进行比較,当某一个列表的元素比較小的话。就将其增加到输出列表中。并将该列表的指针指向列表的下一个元素。这道题是比較简单的,可是有一个边界条件要注意,就是两个列表可能会出现为空的情况,假设l1为空时,能够直接将l2进行返回;假设l2为空时,能够直接将l1返回,这样能够降低非常多计算量。



三.实现代码

#include <iostream>

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/ struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
}; class Solution
{
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
if (!l1)
{
return l2;
} if (!l2)
{
return l1;
} ListNode Head(0);
ListNode *Pre = &Head; while(l1 && l2)
{
if (l1->val < l2->val)
{
Pre->next = l1;
l1 = l1->next;
Pre = Pre->next;
}
else
{
Pre->next = l2;
l2 = l2->next;
Pre = Pre->next;
}
} while (l1)
{
Pre->next = l1;
l1 = l1->next;
Pre = Pre->next;
} while(l2)
{
Pre->next = l2;
l2 = l2->next;
Pre = Pre->next;
} return Head.next; }
};



四.体会

   这道题主要考察的就是边界条件,主要就是处理链表为空的情况,也就是,假设l1为空。就返回l2,假设l2为空,就直接返回l1。

简单的题要考虑充分啊。




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