Skip to content

Latest commit

 

History

History
190 lines (154 loc) · 4.47 KB

File metadata and controls

190 lines (154 loc) · 4.47 KB

English Version

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 dummy,初始值为 $0$,指向链表头节点 head

然后从链表头节点开始遍历,找出链表最后一个值不等于 $9$ 的节点 target,将 target 的值加 $1$。接着将 target 之后的所有节点值置为 $0$

需要注意的是,如果链表中所有节点值都为 $9$,那么遍历结束后,target 会指向空节点,这时我们需要将 dummy 的值加 $1$,然后返回 dummy,否则返回 dummy 的下一个节点。

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

Python3

# 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: ListNode) -> 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

Java

/**
 * 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;
    }
}

C++

/**
 * 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;
        while (head) {
            if (head->val != 9) target = head;
            head = head->next;
        }
        ++target->val;
        target = target->next;
        while (target) {
            target->val = 0;
            target = target->next;
        }
        return dummy->val == 1 ? dummy : dummy->next;
    }
};

Go

/**
 * 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++
	target = target.Next
	for target != nil {
		target.Val = 0
		target = target.Next
	}
	if dummy.Val == 1 {
		return dummy
	}
	return dummy.Next
}

...