LeetCode21. Merge Two Sorted Lists
题目:
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
题目分析:排序合并链表。用递归去实现。
代码:
/**
* 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;
if (l1->val > l2->val) {
ListNode* temp = l2;
temp->next = mergeTwoLists(l1, l2->next);
return temp;
}
else {
ListNode* temp = l1;
temp->next = mergeTwoLists(l1->next, l2);
return temp;
}
}
};
本文介绍了如何使用递归方法来解决LeetCode上的第21题——合并两个有序链表。通过比较两个链表节点的值,选择较小者构建新的链表,最终实现两链表的有序合并。
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