Add Two Numbers

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

以前做过，复杂度太高

题目大意

用链表逆序表示一个非负整数，如342链表表示为2->4->3。两个非负整数链表表示形式相加求和。返回结果链表

思路

1.如果其中一个为空，直接返回不为空的链表

2.两个节点相加，结果放到第一个链表中，保存进位。一直到其中一个或者两个链表结束

3.如果第二个链表不为空，将剩余节点放到第一个链表结尾。对第一个链表进行进位处理

4.如果最后还有进位，申请一个节点，存放进位

时间复杂度为O(N)空间复杂度为O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * l1作为结果，不额外申请空间
     * @param l1
     * @param l2
     * @return
     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       if(l1 == null || l2 == null)                    
           return l1 == null ? l2 : l1;
       ListNode p1 = l1;                            
       ListNode p2 = l2;
       int carry = 0;                                    //记录进位
       ListNode pre = null;
       pre = l1;
       
       //结果存放到p1中
       while(p1 != null && p2 != null){
           p1.val = p2.val + p1.val + carry;
           carry = p1.val / 10;
           p1.val %= 10;
           pre = p1;
           
           p1 = p1.next;
           p2 = p2.next;
       }//while
       
       if(p2 != null){
           p1 = pre;
           p1.next = p2;
           p1 = p1.next;
       }//l2剩余节点放到l1
       
       //对剩余节点进位处理
       while(p1 != null){
           p1.val += carry;
           carry = p1.val / 10;
           p1.val %= 10;
           pre = p1;
           p1 = p1.next;
       }
       
       p1 = pre;
       if(carry == 1){
           ListNode node = new ListNode(carry);
           p1.next = node;
       }
       
       return l1;
    }
}