LeetCode #369. Plus One Linked List

最新推荐文章于 2023-01-05 00:00:32 发布

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本文深入探讨了链表加一算法的两种实现方法。一种是通过翻转链表进行加一操作，再翻转回来；另一种是遍历链表找到最右侧不为9的节点并加一，将右侧节点置零。文章提供了详细的代码实现，适合对链表操作和算法优化感兴趣的读者。

题目描述：

Given a non-negative integer represented as non-empty a singly linked list of digits, plus one to the integer.

You may assume the integer do not contain any leading zero, except the number 0 itself.

The digits are stored such that the most significant digit is at the head of the list.

Example :

Input: [1,2,3]
Output: [1,2,4]

class Solution {
public:
    ListNode* plusOne(ListNode* head) {
        if(head==NULL) return NULL;
        ListNode* rhead=reverse(head);
        int carry=1;
        ListNode* p=rhead;
        ListNode* pre=p;
        while(carry&&p)
        {
            int x=p->val+carry;
            p->val=x%10;
            carry=x/10;
            pre=p;
            p=p->next;
        }
        if(carry) pre->next=new ListNode(1);
        return reverse(rhead);
    }
    
    ListNode* reverse(ListNode* head)
    {
        if(head==NULL||head->next==NULL) return head;
        ListNode* dummy=new ListNode(0);
        dummy->next=head;
        ListNode* p=head;
        ListNode* pre=dummy;
        while(p->next)
        {
            ListNode* tmp=p->next;
            p->next=p->next->next;
            tmp->next=dummy->next;
            dummy->next=tmp;
        }
        return dummy->next;
    }
};

class Solution {
public:
    ListNode* plusOne(ListNode* head) {
        ListNode* p=head;
        ListNode* right=NULL;
        while(p) // 找到最右边不为9的节点标记为right
        {
            if(p->val!=9) right=p;
            p=p->next;
        }
        if(!right) //所有节点都是9的情况下，增加一个节点放在最左边
        {
            right=new ListNode(0);
            right->next=head;
            head=right;
        }
        right->val++;
        p=right->next;
        while(p) //right右边的所有数字变成0
        {
            p->val=0;
            p=p->next;
        }
        return head;
    }
};