21. Merge Two Sorted Lists
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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
解題思路:
/*
執行用時 : 8 ms, 在Merge Two Sorted Lists的C送出中擊敗了97.49% 的使用者
記憶體消耗 : 7.2 MB, 在Merge Two Sorted Lists的C送出中擊敗了93.61% 的使用者
*/
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
//考慮特殊情況
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
//定義傳回結果連結清單,并配置設定空間
struct ListNode *result = (struct ListNode*)malloc(sizeof(struct ListNode));
//始終指向結果連結清單尾,進行插入操作
struct ListNode *temp = result;
//開始插入
while(l1&&l2){
if(l1->val <= l2->val)
{
temp->next = l1;
l1 = l1->next;
}
else
{
temp->next = l2;
l2 = l2->next;
}
//指向結果連結清單尾
temp = temp->next;
}
//還有連結清單沒有合并,連結到結果連結清單尾
temp->next = (l1 != NULL)?l1:l2;
return result->next;
free(result);
}
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if(l1==NULL)
return l2;
if(l2==NULL)
return l1;
if(l1->val<l2->val){
l1->next = mergeTwoLists(l1->next,l2);
return l1;
}
else
{
l2->next = mergeTwoLists(l1,l2->next);
return l2;
}
}
後記:
試想一下對于無序連結清單該如何處理。