[LeetCode] 021. Merge Two Sorted Lists (Easy) (C++/Python)

发布时间：2020-12-13 20:16:08 所属栏目：PHP教程来源：网络整理

导读：索引：[LeetCode] Leetcode 题解索引 (C/Java/Python/Sql) Github: https://github.com/illuz/leetcode 021.Merge_Two_Sorted_Lists (Easy) 链接：题目：https://oj.leetcode.com/problems/merge-two-sorted-lists/ 代码(github)：https://github.com/illu

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

021.Merge_Two_Sorted_Lists (Easy)

链接：

题目：https://oj.leetcode.com/problems/merge-two-sorted-lists/
代码(github)：https://github.com/illuz/leetcode

题意：

合并两个有序链表。

分析：

很经典的题目，不过知道怎样做后很容易，摹拟便可。
有两种做法：
1. 开1个节点做 head 的前节点（下面的 Python 代码实现）
2. 不开直接做（C++ 代码实现）

代码：

C++:

class Solution { public: ListNode *mergeTwoLists(ListNode *l1,ListNode *l2) { if (l1 == NULL) return l2; if (l2 == NULL) return l1; ListNode *start,*cur; if (l1->val < l2->val) { cur = start = l1; l1 = l1->next; } else { cur = start = l2; l2 = l2->next; } while (l1 != NULL && l2 != NULL) { if (l1->val < l2->val) { cur->next = l1; cur = l1; l1 = l1->next; } else { cur->next = l2; cur = l2; l2 = l2->next; } } if (l1 != NULL) cur->next = l1; else cur->next = l2; return start; } }; ListNode *l1,*l2,*ll1,*ll2; int main() { int n1,n2; Solution s; cin >> n1; ll1 = l1 = new ListNode(0); for (int i = 0; i < n1; i++) { l1->next = new ListNode(0); l1 = l1->next; scanf("%d",&(l1->val)); } cin >> n2; ll2 = l2 = new ListNode(0); for (int i = 0; i < n2; i++) { l2->next = new ListNode(0); l2 = l2->next; scanf("%d",&(l2->val)); } ListNode *res = s.mergeTwoLists(ll1->next,ll2->next); while (res != NULL) { cout << res->val << ' '; res = res->next; } return 0; }

Python:

class Solution: # @param two ListNodes # @return a ListNode def mergeTwoLists(self,l1,l2): if not l1 and not l2: return None dummy = ListNode(0) cur = dummy while l1 and l2: if l1.val <= l2.val: cur.next = l1 l1 = l1.next else: cur.next = l2 l2 = l2.next cur = cur.next cur.next = l1 or l2 return dummy.next

（编辑：李大同）

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