LeetCode 369. Plus One Linked List

原题链接在这里：https://leetcode.com/problems/plus-one-linked-list/

题目：

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

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

题解：

Method 1:

末位加一，若是有进位，就要在 Linked List 的前一个 Node 做改动，自然而然想到先Reverse Linked List. 从头开始做比较方便. 加完再reverse回来.

Time Complexity: O(n). reverse 用O(n), reverse 两次 加上 iterate 一次.

Space: O(1).

AC Java:

/**
 * 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 tail = reverse(head);
        ListNode cur = tail;
        ListNode pre = cur;
        int carry = 1;
        
        while(cur != null){
            pre = cur;
            int sum = cur.val + carry;
            cur.val = sum%10;
            carry = sum/10;
            if(carry == 0){
                break;
            }
            cur = cur.next;
        }
        if(carry == 1){
            pre.next = new ListNode(1);
        }
        return reverse(tail);
    }
    
    private ListNode reverse(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        ListNode tail = head;
        ListNode cur = head;
        ListNode pre;
        ListNode temp;
        
        while(tail.next != null){
            pre = cur;
            cur = tail.next;
            temp = cur.next;
            cur.next = pre;
            tail.next = temp;
        }
        return cur;
    }
}

Method 2:

利用递归，因为递归的终止条件就是到了末尾节点，每层向上返回一个carry数值.

Time Complexity: O(n), 最多每个节点iterate两遍.

Space: O(n), recursion用了n层stack.

AC Java:

/**
 * 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;
        }
        int carry = dfs(head);
        if(carry == 1){
            ListNode dummy = new ListNode(1);
            dummy.next = head;
            return dummy;
        }
        return head;
    }
    
    private int dfs(ListNode head){
        if(head == null){
            return 1;
        }
        int carry = dfs(head.next);
        int sum = head.val + carry;
        head.val = sum%10;
        return sum/10;
    }
}

Method 3:

和Method 2原题相同，用stack代替recursion.

Time Complexity: O(n).

Space: O(n). Stack 大小.

/**
 * 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;
        }
        Stack<ListNode> stk = new Stack<ListNode>();
        ListNode cur = head;
        while(cur != null){
            stk.push(cur);
            cur = cur.next;
        }
        int carry = 1;
        while(!stk.isEmpty() && carry == 1){
            ListNode top = stk.pop();
            int sum = top.val + carry;
            top.val = sum%10;
            carry = sum/10;
        }
        if(carry == 1){
            ListNode dummy = new ListNode(1);
            dummy.next = head;
            return dummy;
        }
        return head;
    }
}

Method 4:

从右向左寻找第一个不是9的节点，找到后在该节点加一,  若是他后面还有节点, 说明后面的节点都是9, 所以都要变成0.

Time Complexity: O(n).

Space: O(1).

AC Java:

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        ListNode notNine = dummy;;
        while(cur != null){
            if(cur.val != 9){
                notNine = cur;
            }
            cur = cur.next;
        }
        
        notNine.val += 1;
        cur = notNine.next;
        while(cur != null){
            cur.val = 0;
            cur = cur.next;
        }
        return notNine == dummy ? dummy : head;
    }
}