剑指offer25合并两个排序的列表

Posted 2020-12-05 Heisenber9

题目

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

思路

思路很朴素，谁小先接上谁。

给出递归和非递归两个版本

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        // 递归版本
        if (!pHead1) return pHead2;
        if (!pHead2) return pHead1;
        
        ListNode* mergeList = nullptr;
        
        if (pHead1->val >= pHead2->val){
            mergeList = pHead2;
            mergeList->next = Merge(pHead1, pHead2->next);
        }
        else {
            mergeList = pHead1;
            mergeList->next = Merge(pHead1->next, pHead2);
        }
        return mergeList;
    }
};

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        //  非递归版本
        ListNode* dummy = new ListNode(-1);
        ListNode* res = dummy;
         //不同判断p1和p2是否存在，已经处理
        while (pHead1 && pHead2){ 
            if (pHead1->val >= pHead2->val)
            {
                dummy->next = pHead2;
                pHead2 = pHead2->next;
            }
            else{
                dummy->next = pHead1;
                pHead1 = pHead1->next;
            }
            dummy = dummy->next;
        }
        dummy->next = pHead1?pHead1:pHead2;
        return res->next;
    }
};