leetcode:21 Merge Two Sorted Lists－每日编程第十七题

Merge Two Sorted Lists

Total Accepted: 95133 Total Submissions: 281996 Difficulty: Easy

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

思路：

1).轮流比较l1与l2的值，较小的值链接到新链上。如l1->val<=l2->val,将l1链接到新链上，然后递增l1。

2).重复1).直到l1或者l2为空。

3).最后将非空的l1或者l2整个链接到新链末端，结束。

递归：（递归的方法简单，且比较好理解）

/** * 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) {        if(l1==NULL){            return l2;        }else if(l2==NULL){            return l1;        }else{            if(l1->val<=l2->val){                l1->next=mergeTwoLists(l1->next,l2);                return l1;            }else{                l2->next=mergeTwoLists(l1,l2->next);                return l2;            }        }    }};

非递归：

/** * 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) {        if(l1==NULL){            return l2;        }else if(l2==NULL){            return l1;        }        ListNode* head;        if(l1->val<=l2->val){            head=l1;            l1=l1->next;        }else{            head=l2;            l2=l2->next;        }        ListNode* p=head;        while(l1!=NULL&&l2!=NULL){            if(l1->val<=l2->val){                p->next=l1;                p=l1;                l1=l1->next;            }else{                p->next=l2;                p=l2;                l2=l2->next;            }        }        if(l1==NULL){            p->next=l2;        }        if(l2==NULL){            p->next=l1;        }        return head;    }};

递归：