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;
     }
 }