題目

You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

你有兩個(gè)用鏈表代表的整數(shù)，其中每個(gè)節(jié)點(diǎn)包含一個(gè)數(shù)字。數(shù)字存儲(chǔ)按照在原來整數(shù)中相反的順序，使得第一個(gè)數(shù)字位于鏈表的開頭。寫出一個(gè)函數(shù)將兩個(gè)整數(shù)相加，用鏈表形式返回和。
樣例
給出兩個(gè)鏈表3->1->5->null 和 5->9->2->null，返回8->0->8->null

分析

這道題類似之前的二進(jìn)制求和，只不過換到了鏈表這種數(shù)據(jù)結(jié)構(gòu)之上，同時(shí)二進(jìn)制變成了十進(jìn)制，要考慮好進(jìn)位的計(jì)算。

代碼

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;      
 *     }
 * }
 */
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        // write your code here
        // write your code here
        // save the pre node
        ListNode pre = new ListNode(0);
        // save the now node
        ListNode now = new ListNode(0);
        // create a null node to store the result
        ListNode result = null;
        int val = 0; //
        int add = 0; //jinwei
        
        while( l1 != null || l2 != null )
        {
            val = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) % 10;
            add = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) / 10;
            l1 = (l1 == null ? null : l1.next);
            l2 = (l2 == null ? null : l2.next);
            now.val = val;
            if(result == null)
            {
                result = now;
            }
            pre = now;
            now = new ListNode(0);
            pre.next = now;
        }
        //最后還要多來一次判斷，因?yàn)橛幸环N可能，兩個(gè)鏈表一樣長，最后一位又向上進(jìn)了一位
        val = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) % 10;
        add = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) / 10;
        now.val = val;
        //如果最后一位又向上進(jìn)了一位，新的最后一位不為0，應(yīng)該保留，否則就為0，應(yīng)當(dāng)舍棄
        if(now.val == 0){
            pre.next = null;
        }
        return result;
    }
}