369. Plus One Linked List

发表于 | 分类于

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:

1
2
3
4
5
Input:
1->2->3
Output:
1->2->4

解法1：

考察linkedlist基本功的一道题，先reverse再加1再reverse
C++

1

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode plusOne(ListNode head) {
        if (head == null) {
            return head;
        }
        ListNode headReversed = reverse(head);
        int carry = 1;
        head = headReversed;
        ListNode prev = head;
        while (head != null) {
            prev = head;
            int digit = (head.val + carry) % 10;
            carry = (head.val + carry) / 10;
            head.val = digit;
            head = head.next;
        }
        if (carry != 0) {
            ListNode temp = new ListNode(carry);
            prev.next = temp;
        }
        // reverse back to get the desired result
        return reverse(headReversed);
    }
    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
}

解法2: Recursion

此题也可以用递归的办法做。因为递归会自然的把list倒过来处理。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode plusOne(ListNode head) {
        int carry = dfs(head);
        if (carry == 0) {
            return head;
        }
        ListNode newHead = new ListNode(carry);
        newHead.next = head;
        return newHead;
    }
    private int dfs(ListNode head) {
        if (head == null) return 1;
        int carry = dfs(head.next);
        if (carry == 0) return 0;
        int val = head.val + 1;
        head.val = val % 10;
        return val / 10;
    }
}