leetcode:21 Merge Two Sorted Lists－每日編程第十七題

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).輪流比較l1與l2的值，較小的值鏈接到新鏈上。如l1->val<=l2->val,將l1鏈接到新鏈上，然後遞增l1。

2).重複1).直到l1或者l2爲空。

3).最後將非空的l1或者l2整個鏈接到新鏈末端，結束。

遞歸：（遞歸的方法簡單，且比較好理解）

/**
 * 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;
            }
        }
    }
};

非遞歸：

/**
 * 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;
    }
};

遞歸：