每日一题算法——移除链表元素、反转链表

移除链表元素

力扣题目链接

我的解法：

注意细节：要删掉移除的元素。

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        while(head!=nullptr){
            if(head->val==val){
                head=head->next;
            }
        }
        ListNode* nowhead = head;
        while(nowhead){
            
            if(nowhead->next->val == val){
                if(nowhead->next->next == nullptr){
                    nowhead->next =nullptr;
                }else{
                    nowhead->next=nowhead->next->next;
                }
                
            }
            nowhead = nowhead->next;
        }
        return head;
    }
};

//修改后
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        while(head != NULL && head->val==val){
            ListNode* tmp = head;
            head = head->next;
            delete tmp;
        }
        ListNode* nowhead = head;
        while(nowhead !=NULL && nowhead->next !=NULL ){
            
            if(nowhead->next->val == val){
                    ListNode* tmp=nowhead->next;
                    nowhead->next=nowhead->next->next;
                    delete tmp;
                }else{
                     nowhead = nowhead->next;
                }
                
        }
        return head;
        
    }
};

方法二：

增加一个伪头结点dummyhead，dummyhead->next = head;

这样可以统一头结点和后续节点的删除方式。

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {

        ListNode* dummyhead = new ListNode(0);\\记得定义
        dummyhead->next =head;
        ListNode* nowhead =dummyhead;
        while(nowhead !=NULL && nowhead->next !=NULL ){
            
            if(nowhead->next->val == val){
                    ListNode* tmp=nowhead->next;
                    nowhead->next=nowhead->next->next;
                    delete tmp;
                }else{
                     nowhead = nowhead->next;
                }
                
        }
        return dummyhead->next;
        
    }
};

反转链表

力扣题目链接

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = nullptr;
        ListNode* cur = head;
        while(cur!=nullptr){
            ListNode * tmp = cur->next;
            cur->next =prev;
            prev =cur;
            cur =tmp;
        }
        return prev;
    }
};