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.
思路:
思路很简单,遍历两个链表,每次取最小值。
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) {
ListNode * head = new ListNode(0);
ListNode * ptr = head;while(l1 != NULL || l2 != NULL){ ListNode * cur = NULL; if(l1 == NULL){ cur = l2; l2 = l2->next; } else if(l2 == NULL){ cur = l1; l1 = l1->next; } else{ if(l1->val < l2->val){ cur = l1; l1 = l1->next; } else{ cur = l2; l2 = l2->next; } } ptr->next = cur; ptr = ptr->next; } return head->next;
}
};
任务六(letecode)
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转载自blog.csdn.net/weixin_41741008/article/details/89048380
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