LeetCode Add Two Numbers II

原题链接在这里：https://leetcode.com/problems/add-two-numbers-ii/

题目：

You are given two linked lists representing two non-negative numbers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

题解：

类似Add Two Numbers.

可看成是Add Two Numbers与Reverse Linked List的综合. 先reverse在逐个add, 最后把结果reverse回来.

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; }
  * }
  */
 class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         l1 = reverse(l1);
         l2 = reverse(l2);

         ListNode dummy = new ListNode(0);
         ListNode cur = dummy;
         int carry = 0;

         while(l1 != null || l2!= null){
             if(l1 != null){
                 carry += l1.val;
                 l1 = l1.next;
             }

             if(l2 != null){
                 carry += l2.val;
                 l2 = l2.next;
             }

             cur.next = new ListNode(carry%10);
             carry /= 10;
             cur = cur.next;
         }

         if(carry != 0){
             cur.next = new ListNode(carry);
         }

         ListNode head = dummy.next;
         dummy.next = null;
         return reverse(head);
     }

     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;
     }
 }

也可以使用两个stack把list 1 和 list 2 分别压进去. 再pop出来相加放到new list的head位置.

Time Complexity: O(n). 压stack用了O(n), pop后相加用了O(n).

Space: O(n). stack用了O(n). result list用了O(n).

AC Java:

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         if(l1 == null){
             return l2;
         }
         if(l2 == null){
             return l1;
         }

         Stack<Integer> stk1 = new Stack<Integer>();
         Stack<Integer> stk2 = new Stack<Integer>();
         while(l1 != null){
             stk1.push(l1.val);
             l1 = l1.next;
         }
         while(l2 != null){
             stk2.push(l2.val);
             l2 = l2.next;
         }

         ListNode dummy = new ListNode(0);
         int carry = 0;
         while(!stk1.isEmpty() || !stk2.isEmpty()){
             if(!stk1.isEmpty()){
                 carry += stk1.pop();
             }
             if(!stk2.isEmpty()){
                 carry += stk2.pop();
             }
             ListNode cur = new ListNode(carry%10);
             cur.next = dummy.next;
             dummy.next = cur;
             carry /= 10;
         }
         if(carry != 0){
             ListNode cur = new ListNode(1);
             cur.next = dummy.next;
             dummy.next = cur;
         }
         return dummy.next;
     }
 }