leetcode 之 Merge Two Sorted Lists
2017-09-23 11:06
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题目如下:
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.
分析:
两个已经排好序的链表,要将他们合并成为一个新的有序链表,我们可以遍历这两个链表,比较他们表头的大小,声明一个新的链表,每次把二者较小的那个加入到新的链表中。如果某一个链表到达了末尾,那就把另外一个链表的所有部分加到新链表之后,这样只需一次遍历即可。
代码实现如下:
另外还可以利用递归的方法:
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.
分析:
两个已经排好序的链表,要将他们合并成为一个新的有序链表,我们可以遍历这两个链表,比较他们表头的大小,声明一个新的链表,每次把二者较小的那个加入到新的链表中。如果某一个链表到达了末尾,那就把另外一个链表的所有部分加到新链表之后,这样只需一次遍历即可。
代码实现如下:
/** * 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) { ListNode temp(0); ListNode * t = &temp; while(l1 != NULL && l2 != NULL) { if(l1->val < l2->val) { t->next = l1; l1 = l1->next; } else { t->next = l2; l2 = l2->next; } t = t->next; } if(l1 == NULL) t->next = l2; if(l2 == NULL) t->next = l1; return temp.next; } };
另外还可以利用递归的方法:
class Solution { public: ListNode *mergeTwoLists(ListNode *l1, 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(l2->next, l1); return l2; } } };
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