LeetCode 445. 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;
    }
}