链表合并

题目描述

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。

题目链接：牛客网

模拟归并排序的合并的过程～
可以用迭代，也可以用递归，递归的写法十分整洁易读！！！

// 递归版本/*struct ListNode {    int val;    struct ListNode *next;    ListNode(int x) :            val(x), next(NULL) {    }};*/class Solution {public:    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)    {        if (pHead1 == NULL)            return pHead2;        if (pHead2 == NULL)            return pHead1;        ListNode* head;        if (pHead1->val <= pHead2->val) {            head = pHead1;            head->next = Merge(pHead1->next, pHead2);        } else {            head = pHead2;            head->next = Merge(pHead1, pHead2->next);        }        return head;    }};

// 迭代版本/*struct ListNode {    int val;    struct ListNode *next;    ListNode(int x) :            val(x), next(NULL) {    }};*/class Solution {public:    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {        if (pHead1 == NULL)            return pHead2;        if (pHead2 == NULL)            return pHead1;        // 找出第一个节点        ListNode* now = NULL;        if (pHead1->val <= pHead2->val) {            now = pHead1;            pHead1 = pHead1->next;        } else {            now = pHead2;            pHead2 = pHead2->next;        }        ListNode* head = now;        // 模仿归并排序的合并操作        while (pHead1 != NULL || pHead2 != NULL) {            if (pHead1 == NULL) {                MergeHelper(now, pHead2);                continue;            }            if (pHead2 == NULL) {                MergeHelper(now, pHead1);                continue;            }            if (pHead1->val <= pHead2->val)                MergeHelper(now, pHead1);            else                MergeHelper(now, pHead2);        }        return head;    }    void MergeHelper(ListNode*& master, ListNode*& guest) {        master->next = guest;        master = master->next;        guest = guest->next;    }};