标签:leetcode 算法 java python c++
索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github:
https://github.com/illuz/leetcode
题目:https://oj.leetcode.com/problems/merge-two-sorted-lists/
代码(github):https://github.com/illuz/leetcode
合并两个有序链表。
很经典的题目,不过知道怎么做后很容易,模拟即可。
有两种做法:
1. 开一个节点做 head 的前节点 (下面的 Python 代码实现)
2. 不开直接做(C++ 代码实现)
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
[LeetCode] 021. Merge Two Sorted Lists (Easy) (C++/Python)
标签:leetcode 算法 java python c++
原文地址:http://blog.csdn.net/hcbbt/article/details/44064639