369. 给单链表加一 🔒

题目描述

给定一个用链表表示的非负整数， 然后将这个整数 再加上 1 。

这些数字的存储是这样的：最高位有效的数字位于链表的首位 head 。

 

示例 1:

输入: head = [1,2,3]
输出: [1,2,4]

示例 2:

输入: head = [0]
输出: [1]

 

提示：

• 链表中的节点数在 [1, 100] 的范围内。
• 0 <= Node.val <= 9
• 由链表表示的数字不包含前导零，除了零本身。

解法

方法一：链表遍历

我们先设置一个虚拟头节点 \(\textit{dummy}\)，初始时 \(\textit{dummy}\) 的值为 \(0\)，并且 \(\textit{dummy}\) 的后继节点为链表 \(\textit{head}\)。

接下来，我们从虚拟头节点开始遍历链表，找到最后一个不为 \(9\) 的节点，将其值加 \(1\)，并将该节点之后的所有节点的值置为 \(0\)。

最后，我们判断虚拟头节点的值是否为 \(1\)，如果为 \(1\)，则返回 \(\textit{dummy}\)，否则返回 \(\textit{dummy}\) 的后继节点。

时间复杂度 \(O(n)\)，其中 \(n\) 是链表的长度。空间复杂度 \(O(1)\)。

Python3JavaC++GoTypeScript

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def plusOne(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        target = dummy
        while head:
            if head.val != 9:
                target = head
            head = head.next
        target.val += 1
        target = target.next
        while target:
            target.val = 0
            target = target.next
        return dummy if dummy.val else dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode plusOne(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode target = dummy;
        while (head != null) {
            if (head.val != 9) {
                target = head;
            }
            head = head.next;
        }
        ++target.val;
        target = target.next;
        while (target != null) {
            target.val = 0;
            target = target.next;
        }
        return dummy.val == 1 ? dummy : dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* plusOne(ListNode* head) {
        ListNode* dummy = new ListNode(0, head);
        ListNode* target = dummy;
        for (; head; head = head->next) {
            if (head->val != 9) {
                target = head;
            }
        }
        target->val++;
        for (target = target->next; target; target = target->next) {
            target->val = 0;
        }
        return dummy->val ? dummy : dummy->next;
    }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func plusOne(head *ListNode) *ListNode {
    dummy := &ListNode{0, head}
    target := dummy
    for head != nil {
        if head.Val != 9 {
            target = head
        }
        head = head.Next
    }
    target.Val++
    for target = target.Next; target != nil; target = target.Next {
        target.Val = 0
    }
    if dummy.Val == 1 {
        return dummy
    }
    return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function plusOne(head: ListNode | null): ListNode | null {
    const dummy = new ListNode(0, head);
    let target = dummy;
    while (head) {
        if (head.val !== 9) {
            target = head;
        }
        head = head.next;
    }
    target.val++;
    for (target = target.next; target; target = target.next) {
        target.val = 0;
    }
    return dummy.val ? dummy : dummy.next;
}

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