Merge Two Sorted Lists——解題報告

標籤：leetcode   鏈表   合并   遞迴   指標   

    【題目】

    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.

    【分析】

    不要忘記先判斷兩個鏈表是否有空鏈表。其餘的使用遞迴和非遞迴方式，都可以實現。

    【代碼】

    遞迴方式：

class Solution {public:    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        ListNode* mergedList = NULL;                 if(l1 == NULL)            return l2;        if(l2 == NULL)            return l1;                if(l1->val < l2->val)        {            mergedList = l1;            mergedList->next = mergeTwoLists(l1->next, l2);        }        else        {            mergedList = l2;            mergedList->next = mergeTwoLists(l1, l2->next);        }                return mergedList;            }};

    非遞迴方式：

class Solution {public:    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        ListNode* mergedList = NULL;                 if(l1 == NULL)            return l2;        if(l2 == NULL)            return l1;                if(l1->val < l2->val)        {            mergedList = l1;            mergedList->next = NULL;            l1 = l1->next;        }        else        {            mergedList = l2;             mergedList->next = NULL;            l2 = l2->next;        }                ListNode* p = mergedList;        while(l1 != NULL && l2 != NULL)        {            if(l1->val < l2->val)            {                p->next = l1;                l1 = l1->next;                p = p->next;                p->next = NULL;            }            else            {                p->next = l2;                 l2 = l2->next;                p = p->next;                p->next = NULL;            }        }                if(l1 != NULL)            p->next = l1;         else if(l2 != NULL)            p->next = l2;                return mergedList;            }};

Merge Two Sorted Lists——解題報告