合并单链表

作者: 飞白非白 | 来源:发表于2018-12-04 23:35 被阅读9次
    • 合并两个有序链表非递归实现
    /**
     * 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* header = new ListNode(-1);
            ListNode* cur = header;
            while(l1 || l2)
            {
                int n = 0;
                if(l1 == NULL || (l1 && l2 && l1->val > l2->val))
                {
                    n = l2->val;
                    l2 = l2->next;
                }
                else if(l2 == NULL || (l1 && l2 && l1->val <= l2->val))
                {
                    n = l1->val;
                    l1 = l1->next;
                }
    
                cur->next = new ListNode(n);
                cur = cur->next;
            }
    
            ListNode *ans = header->next;
            delete header;
            return ans;
        }
    };
    
    • 合并两个有序链表递归实现
    Node* MergeListR(Node* Head1,Node* Head2)
    {
        if(NULL == Head1 || NULL == Head2){
            if(NULL == Head1)
                return Head2;
    
            return Head1;
        }
    
        Node* newHead = NULL;
        //较小的作为新链表的头
        if(Head1->_data < Head2->_data){
            newHead = Head1;
            newHead->_next = MergeList(newHead->_next,Head2);
        }else{
            newHead = Head2;
            newHead->_next = MergeList(Head1,newHead->_next);
        }
    
        return newHead;
    }
    

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