add two numbers

1. Question

给两个链表，分别用于表示两个非负数字。数的每一位在链表中反序存放（e.g. 2->4->3代表数342），且每个节点一位。对这两个数求和并返回，结果也用链表表示。

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

2. Solution

定义四个变量：

1. i：数1位指针，初值为链表头
2. j：数2位指针，初值为里边头
3. k：进位，初值为0
4. res：结果链表指针

*res = ( *i + *j + k ) %10;

k = ( *i + *j + k ) / 10;

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if( l1 == null ) return l2;
        if( l2 == null ) return l1;
        
        int quotient = 0;
        ListNode p = new ListNode(0);
        p.next = l1;    //use l1 as the result link list
        ListNode q = l2;
        while( p.next!=null && q!=null ){
            p.next.val = p.next.val + q.val + quotient;
            quotient = p.next.val / 10;
            p.next.val %= 10;
            p = p.next;
            q = q.next;
        }
        if( q != null ) p.next = q;
        while( p.next != null && quotient != 0 ){
            p.next.val += quotient;
            quotient = p.next.val / 10;
            p.next.val %= 10;
            p = p.next;
        }
        if( quotient != 0 )
            p.next = new ListNode(quotient);
        
        return l1;
    }
}

3. 复杂度分析

算法遍历链表，时间复杂度O(n)