21. Merge Two Sorted Lists【easy】
21. Merge Two Sorted Lists【easy】
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.
解法一:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { 12 if (l1 == NULL || l2 == NULL) { 13 return l1 ? l1 : l2; 14 } 15 16 ListNode * dummy = new ListNode(INT_MIN); 17 ListNode * temp = dummy; 18 19 while (l1 && l2) { 20 if (l1->val > l2->val) { 21 dummy->next = l2; 22 l2 = l2->next; 23 } 24 else { 25 dummy->next = l1; 26 l1 = l1->next; 27 } 28 29 dummy = dummy->next; 30 } 31 32 if (l1 || l2) { 33 dummy->next = l1 ? l1 : l2; 34 } 35 36 return temp->next; 37 } 38 };
由于最后是弄到list1中,但是我们不知道list1还是list2的第一个元素关系,最后结果的list1中的头结点可能会改变,所以需要引入dummy节点。
解法二:
1 public ListNode mergeTwoLists(ListNode l1, ListNode l2){ 2 if(l1 == null) return l2; 3 if(l2 == null) return l1; 4 if(l1.val < l2.val){ 5 l1.next = mergeTwoLists(l1.next, l2); 6 return l1; 7 } else{ 8 l2.next = mergeTwoLists(l1, l2.next); 9 return l2; 10 } 11 }
参考了@yangliguang 的代码
解法三:
1 public class Solution { 2 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 3 if (l1 == null) return l2; 4 if (l2 == null) return l1; 5 6 ListNode handler; 7 if(l1.val < l2.val) { 8 handler = l1; 9 handler.next = mergeTwoLists(l1.next, l2); 10 } else { 11 handler = l2; 12 handler.next = mergeTwoLists(l1, l2.next); 13 } 14 15 return handler; 16 } 17 }
参考了@RunRunCode 的代码
解法二和解法三都是递归,还没有完全弄明白……
posted on 2017-10-14 13:06 LastBattle 阅读(158) 评论(0) 编辑 收藏 举报