LeetCode 2. Add Two Numbers

原题链接：https://leetcode.com/problems/add-two-numbers/

题目：

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

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

题解：

思路两个val和进位carry相加，组成新点往后连，注意两个list长度不同和最后是否还有一个进位的情况。

Time Complexity: O(n), n是较长list的长度.

Space: 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;
        }
        
        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);
            cur = cur.next;
            carry = carry/10;
        }
        if(carry != 0){
            cur.next = new ListNode(carry);
        }
        return dummy.next;
    }
}

也可以选择更改较长list的node val.

不需要n的extra space.

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 addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        int len1 = 0;
        ListNode cur = l1;
        while(cur != null){
            cur = cur.next;
            len1++;
        }
        
        int len2 = 0;
        cur = l2;
        while(cur != null){
            cur = cur.next;
            len2++;
        }
        
        ListNode dummy = new ListNode(0);
        if(len1 > len2){
            dummy.next = l1;
        }else{
            dummy.next = l2;
        }
        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.val = carry%10;
            cur = cur.next;
            carry /= 10;
        }
        if(carry != 0){
            cur.next = new ListNode(1);
        }
        return dummy.next;
    }
}

跟上Add Two Numbers II, Sum of Two Integers.