Leetcode Plus One Linked List

The original title link is here: https://leetcode.com/problems/plus-one-linked-list/

Topic:

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

The digits is stored such, the most significant digit was at the head of the list.

Example:

Input:1->2->3output:1->2->4

Exercises

Method 1:

The last plus one, if there is carry, it is necessary to change the Linked list before a Node, naturally think of reverse Linked list first. It's easier to do it from the beginning. Add up and reverse back.

Time Complexity:o (n). Reverse with O (n), reverse two times plus iterate once.

Space:o (1).

AC Java:

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7  * }8  */9  Public classSolution {Ten      PublicListNode PlusOne (ListNode head) { One         if(Head = =NULL){ A             returnhead; -         } -ListNode tail =reverse (head); theListNode cur =tail; -ListNode pre =cur; -         intCarry = 1; -          +          while(cur! =NULL){ -Pre =cur; +             intsum = Cur.val +carry; ACur.val = sum%10; atcarry = SUM/10; -             if(Carry = = 0){ -                  Break; -             } -Cur =Cur.next; -         } in         if(Carry = = 1){ -Pre.next =NewListNode (1); to         } +         returnreverse (tail); -     } the      *     PrivateListNode Reverse (listnode head) { $         if(Head = =NULL|| Head.next = =NULL){Panax Notoginseng             returnhead; -         } theListNode tail =head; +ListNode cur =head; A ListNode pre; the ListNode temp; +          -          while(Tail.next! =NULL){ $Pre =cur; $Cur =Tail.next; -temp =Cur.next; -Cur.next =Pre; theTail.next =temp; -         }Wuyi         returncur; the     } -}

Method 2:

Recursion is used because the terminating condition of the recursion is to the end node, and each layer returns a carry value.

Time Complexity:o (n), up to a maximum of two times per node iterate.

Space:o (n), recursion uses the N-layer stack.

AC Java:

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7  * }8  */9  Public classSolution {Ten      PublicListNode PlusOne (ListNode head) { One         if(Head = =NULL){ A             returnhead; -         } -         intcarry =Dfs (head); the         if(Carry = = 1){ -ListNode dummy =NewListNode (1); -Dummy.next =head; -             returndummy; +         } -         returnhead; +     } A      at     Private intDfs (ListNode head) { -         if(Head = =NULL){ -             return1; -         } -         intcarry =DFS (head.next); -         intsum = Head.val +carry; inHead.val = sum%10; -         returnSum/10; to     } +}

Method 3:

Same as Method 2, use stack instead of recursion.

Time Complexity:o (n).

Space:o (n). Stack size.

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7  * }8  */9  Public classSolution {Ten      PublicListNode PlusOne (ListNode head) { One         if(Head = =NULL){ A             returnhead; -         } -stack<listnode> STK =NewStack<listnode>(); theListNode cur =head; -          while(cur! =NULL){ - Stk.push (cur); -Cur =Cur.next; +         } -         intCarry = 1; +          while(!stk.isempty () && carry = = 1){ AListNode top =Stk.pop (); at             intsum = Top.val +carry; -Top.val = sum%10; -carry = SUM/10; -         } -         if(Carry = = 1){ -ListNode dummy =NewListNode (1); inDummy.next =head; -             returndummy; to         } +         returnhead; -     } the}

Method 4:

From the right to the left to find the first node is not 9, found after the node plus a, if he has a node behind, indicating that the following nodes are 9, so all to become 0.

Time Complexity:o (n).

Space:o (1).

AC Java:

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7  * }8  */9  Public classSolution {Ten      PublicListNode PlusOne (ListNode head) { One         if(Head = =NULL){ A             returnhead; -         } -stack<listnode> STK =NewStack<listnode>(); theListNode cur =head; -          while(cur! =NULL){ - Stk.push (cur); -Cur =Cur.next; +         } -         intCarry = 1; +          while(!stk.isempty () && carry = = 1){ AListNode top =Stk.pop (); at             intsum = Top.val +carry; -Top.val = sum%10; -carry = SUM/10; -         } -         if(Carry = = 1){ -ListNode dummy =NewListNode (1); inDummy.next =head; -             returndummy; to         } +         returnhead; -     } the}

Leetcode Plus One Linked List