题目
https://leetcode.com/problems/merge-two-sorted-lists
题意
合并两个有序链表,并且合并结果的链表也是有序的。
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
思路
author's blog == http://www.cnblogs.com/toulanboy/
使用合并排序中的合并思想。
具体做法是:分别用i,j指向两个已排序的链表首个元素。第i个元素和第j个元素比较,谁小谁去新链表。
在我的做法中,我并没有创建新的节点,新链表中的节点都是来自输入数据的2条链表。
代码
//author's blog == http://www.cnblogs.com/toulanboy/
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL)
return l2;
if(l2 == NULL)
return l1;
ListNode * l3 = NULL;
//先处理第1个节点
if(l1->val <= l2->val){
l3 = l1;
l1 = l1->next;
}
else{
l3 = l2;
l2 = l2->next;
}
//让p指向l3的最后一个节点
ListNode * p = l3;
while(l1 && l2){
if(l1->val <= l2->val){
p->next = l1;
p = l1;
l1 = l1->next;
}
else{
p->next = l2;
p = l2;
l2 = l2->next;
}
}
if(l1){
p->next = l1;
}
if(l2){
p->next = l2;
}
return l3;
}
};//author's blog == http://www.cnblogs.com/toulanboy/
运行结果
Runtime: 8 ms, faster than 96.79% of C++ online submissions for Merge Two Sorted Lists.
Memory Usage: 9 MB, less than 65.00% of C++ online submissions for Merge Two Sorted Lists.