提升代码能力第三天

Leecode-21

此题好像不太需要什么算法，还是锻炼链表的熟悉程度，以下是一看到题目马上写的

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode*list = new ListNode(0);
        ListNode*cur = list;
        while(l1||l2){
            if(l2 == NULL){
                cur->next=l1;
                break;
            }
            else if(l1 == NULL){
                cur->next=l2;
                break;
            }
            else if(l1->val <= l2->val){
                cur->next = new ListNode(l1->val);
                l1 = l1->next;
                cur = cur->next;
                
            }
            else {
                cur->next = new ListNode(l2->val);
                l2 = l2->next;
                cur = cur->next;
            }

        }
       return list->next; 
    }
};

以下为递归算法

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode*list = new ListNode(0);
        ListNode*cur = list;
        while(l1||l2){
            if(l2 == NULL){
                return l1;
            }
            else if(l1 == NULL){
                return l2;
            }
            else if(l1->val <= l2->val){
                l1->next = mergeTwoLists(l1->next,l2);
                return l1;
            }
            else {
                l2->next = mergeTwoLists(l1,l2->next);
                return l2;
            }

        }
       return list->next; 
    }
};

以下为迭代算法

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode*prehead = new ListNode(0);
        ListNode*cur = prehead;
        while(l1||l2){
            if(l2 == NULL){
                cur->next=l1;
                break;
            }
            else if(l1 == NULL){
                cur->next=l2;
                break;
            }
            else if(l1->val <= l2->val){
                cur->next = l1;
                l1 = l1->next;
                cur = cur->next;
                
            }
            else {
                cur->next = l2;
                l2 = l2->next;
                cur = cur->next;
            }

        }
       return prehead->next; 
    }
};

迭代算法相较于第一种节省了空间，只是改变指针，没有开辟新的空间，空间复杂度为O（1）
时间复杂度均为0（m+n）
哨兵节点prehead