Add Two Numbers

https://oj.leetcode.com/problems/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

下面是一个不能处理超长数据的答案。

/**
 * 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) {
        ListNode node = l1;
        int l1Value = 0;
        int count = 0;
        while(node != null){
            l1Value = l1Value + node.val * ((Double)Math.pow(10, count)).intValue();
            count ++;
            node = node.next;
        }
        
        node = l2;
        int l2Value = 0;
        count = 0;
        while(node != null){
            l2Value = l2Value + node.val * ((Double)Math.pow(10, count)).intValue();
            count++;
            node = node.next;
        }
        
        int sumValue = l1Value + l2Value;
        
        //int digits = (Math.floor((Double)(Math.log(sumValue) / Math.log(10) + 1)).intValue()); 
        int digits = String.valueOf(sumValue).length();
        if(sumValue < 10){
            digits = 1;
        }
        ListNode[] listNodes = new ListNode[digits];
        
        listNodes[0] = new ListNode(sumValue % 10);
        sumValue = sumValue / 10;
        
        for(int i = 1; i < digits; i++){
            listNodes[i] = new ListNode(sumValue % 10);
            listNodes[i - 1].next = listNodes[i];
            sumValue = sumValue / 10;
        }
        return listNodes[0];
    }
}

 下面是一个AC的方法。技巧是，对于为null的node，全部置为0，再运算。

/**
 * 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) {
        ListNode node1 = l1;
        ListNode node2 = l2;
        ListNode node_current = new ListNode(0);
        ListNode node_next = node_current;
        ListNode node_return = node_current;
        
        while(node1 != null || node2 != null){
            if(node1 == null){
                node1 = new ListNode(0);
            }
            if(node2 == null){
                node2 = new ListNode(0);
            }
            int sum = node1.val + node2.val + node_current.val;
            node_current.val = (node1.val + node2.val + node_current.val) % 10;
            if(sum > 9 || node1.next != null || node2.next !=null){
                node_next = new ListNode(sum / 10);
                node_current.next = node_next;
            }
            node_current = node_next;
            if(node1 != null){
                node1 = node1.next;
            }
            if(node2 != null){
                node2 = node2.next;
            }
        }

        return node_return;
    }
}

 //20180626

/**
 * 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) {
        ListNode head = new ListNode(0);
        ListNode cur = head;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int sum = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = sum / 10;
            sum = sum % 10;
            cur.next = new ListNode(sum);
            cur = cur.next;
            if (l1 != null) {
                l1 = l1.next;
            }
            if (l2 != null) {
                l2 = l2.next;
            }
        }
        if (carry > 0) {
            cur.next = new ListNode(carry);
        }
        return head.next;
    }
}