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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。

思路

很基础的两个指针分别指向两个链表，然后比大小的过程，循环解决。代码写的很细致，主要考虑鲁棒性问题。

代码

/*struct ListNode {	int val;	struct ListNode *next;	ListNode(int x) :			val(x), next(NULL) {	}};*/class Solution {public:    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)    {                ListNode* p1 = pHead1;        ListNode* p2 = pHead2;                ListNode* Head = NULL;        ListNode* pLast = NULL;                if(pHead1 == NULL && pHead2 == NULL){            return NULL;        }else if(pHead1 != NULL && pHead2 == NULL){            return pHead1;        }else if(pHead1 == NULL && pHead2 != NULL){            return pHead2;        }                        if(p1->val < p2->val){            Head = new ListNode(p1->val);            pLast = Head;            p1=p1->next;        }else if(p1->val > p2->val){            Head = new ListNode(p2->val);            pLast = Head;            p2=p2->next;        }else if(p1->val == p2->val){            Head = new ListNode(p1->val);            pLast = Head;            pLast->next = new ListNode(p2->val);            pLast=pLast->next;            p1=p1->next;            p2=p2->next;        }                while(p1 != NULL && p2!= NULL){            if(p1->val < p2->val){                pLast->next = new ListNode(p1->val);                pLast = pLast->next;                p1=p1->next;            }else if(p1->val > p2->val){                pLast->next = new ListNode(p2->val);                pLast = pLast->next;                p2=p2->next;            }else if(p1->val == p2->val){                pLast->next = new ListNode(p1->val);                pLast=pLast->next;                pLast->next = new ListNode(p2->val);                pLast=pLast->next;                p1=p1->next;                p2=p2->next;            }        }                if(p2 != NULL){            while(p2!= NULL){                pLast->next = new ListNode(p2->val);                pLast = pLast->next;                p2=p2->next;            }        }else if(p1 != NULL){            while(p1!= NULL){                pLast->next = new ListNode(p1->val);                pLast = pLast->next;                p1=p1->next;            }        }                return Head;    }};

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