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21. Merge Two Sorted Lists
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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.
题目大意:合并两个有序的链表
思路:通过比较两个链表的节点大小,采用尾插法建立链表。
代码如下:
/** * 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 * newListHead,* newListNode,*newListTail; newListHead = (ListNode *)malloc(sizeof(ListNode)); newListTail = newListHead; while( (NULL != l1) && (NULL != l2) ) { if(l1->val <= l2->val) { newListNode = (ListNode *)malloc(sizeof(ListNode)); newListNode->val = l1->val; newListTail->next = newListNode; newListTail = newListNode; l1 = l1->next; } else { newListNode = (ListNode *)malloc(sizeof(ListNode)); newListNode->val = l2->val; newListTail->next = newListNode; newListTail = newListNode; l2 = l2->next; } } if(NULL != l1) { while(l1) { newListNode = (ListNode *)malloc(sizeof(ListNode)); newListNode->val = l1->val; newListTail->next = newListNode; newListTail = newListNode; l1 = l1->next; } } if(NULL != l2) { while(l2) { newListNode = (ListNode *)malloc(sizeof(ListNode)); newListNode->val = l2->val; newListTail->next = newListNode; newListTail = newListNode; l2 = l2->next; } } newListTail->next = NULL; return newListHead->next; } };
2016-08-06 01:40:31
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