剑指offer系列——16.合并两个排序的链表

Q：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。
T：
1.我犯了个傻，没注意是两个单调递增的链表，误以为是两个无序的链表。

    ListNode *Merge(ListNode *pHead1, ListNode *pHead2) {
        if(pHead1 == nullptr)
            return pHead2;
        if(pHead2 == nullptr)
            return pHead1;
        vector<int> array;
        while (pHead1) {
            array.push_back(pHead1->val);
            pHead1 = pHead1->next;
        }
        while (pHead2) {
            array.push_back(pHead2->val);
            pHead2 = pHead2->next;
        }
        stable_sort(array.begin(), array.end(), [](const int &a,const int &b){return a<b;});
        int length = array.size();
        ListNode *target = new ListNode(array[0]);
        ListNode *L = target;
        for (int i = 1; i < length; i++) {
            ListNode *temp = new ListNode(array[i]);
            target->next = temp;
            target = temp;
        }
        return L;
    }

2.常用的方法

ListNode *Merge(ListNode *pHead1, ListNode *pHead2) {
        int temp = 0;
        if (pHead1 == nullptr)
            return pHead2;
        if (pHead2 == nullptr)
            return pHead1;
        if (pHead1->val <= pHead2->val) {
            temp = pHead1->val;
            pHead1 = pHead1->next;
        } else {
            temp = pHead2->val;
            pHead2 = pHead2->next;

        }
        ListNode* L = new ListNode(temp);
        ListNode *list = L;
        while (pHead1 && pHead2) {
            if (pHead1->val <= pHead2->val) {
                L->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                L->next = pHead2;
                pHead2 = pHead2->next;
            }
            L = L->next;
        }
        if (pHead1 != nullptr)
            L->next = pHead1;
        if (pHead2 != nullptr)
            L->next = pHead2;
        return list;
    }

3.递归

    ListNode *Merge(ListNode *pHead1, ListNode *pHead2){
        if (pHead1 == nullptr)
            return pHead2;
        if (pHead2 == nullptr)
            return pHead1;
        if(pHead1->val<=pHead2->val) {
            pHead1->next = Merge(pHead1->next, pHead2);
            return pHead1;
        }
        else {
            pHead2->next = Merge(pHead1, pHead2->next);
            return pHead2;
        }
    }