[LeetCode] Merge Two Sorted Lists, Solution
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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.
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[Thoughts]
简单的实现,也没什么可说的。
[Code]
1: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
2: if(l1 == NULL) return l2;
3: if(l2 == NULL) return l1;
4: ListNode *head = new ListNode(-1);
5: ListNode *p = head;
6: while(l1 != NULL && l2!=NULL)
7: {
8: if(l1->val < l2->val)
9: {
10: p->next = l1;
11: l1= l1->next;
12: }
13: else
14: {
15: p->next = l2;
16: l2 = l2->next;
17: }
18: p = p->next;
19: }
20: if(l1 != NULL)
21: p->next = l1;
22: if(l2 != NULL)
23: p->next = l2;
24: p = head->next;
25: delete head;
26: return p;
27: }
Update 04/13/13 refactor code for succinct
1: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
2: ListNode* head = new ListNode(-1);
3: ListNode* p = head;
4: while(l1!=NULL || l2!= NULL)
5: {
6: int val1 = l1==NULL?INT_MAX:l1->val;
7: int val2 = l2==NULL? INT_MAX:l2->val;
8: if(val1<=val2)
9: {
10: p->next = l1;
11: l1=l1->next;
12: }
13: else
14: {
15: p->next = l2;
16: l2 = l2->next;
17: }
18: p= p->next;
19: }
20: p = head->next;
21: delete head;
22: return p;
23: }
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