2. Add Two Numbers.md

2. Add Two Numbers

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Solution

1. LinkedList的遍历。
2. 总计的几点技巧：
   • 在遍历的过程中，要注意判断非空的情况。
   • 使用一个dummy node保存链表头。
   • 记得每一次都要将cur结点向后移动一格（此处不做会造成TLE）。
3. 要将能提取出来的公共流程都单独提取出来。

/**
 * 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) {
        int carry = 0;
        ListNode res = new ListNode(0);
        ListNode cur = res;
        while(l1 != null || l2 != null){
            int val = 0;
            if(l1 != null && l2 != null){
                val = l1.val + l2.val + carry;
                l1 = l1.next;
                l2 = l2.next;
            }else if(l1 != null){
                val = l1.val + carry;
                l1 = l1.next;
            }else{
                val = l2.val + carry;
                l2 = l2.next;
            }
            cur.next = new ListNode(val % 10);
            carry = val / 10;
            cur = cur.next;
        }
        if(carry == 1)
            cur.next = new ListNode(1);
        return res.next;
    }
}

Second Time

• Method 1:

   /**
    * 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) {
   		int carry = 0;
   		ListNode cur1 = l1, cur2 = l2;
   		ListNode dummy = new ListNode(0);
   		ListNode cur = dummy;
   		ListNode temp = null;
   		while(cur1 != null && cur2 != null){
   			int sum = cur1.val + cur2.val + carry;
   			temp = new ListNode(sum % 10);
   			carry = sum / 10;
   			cur.next = temp;
   			cur = cur.next;
   			cur1 = cur1.next;
   			cur2 = cur2.next;
   		}
   		ListNode node = cur1 != null ? cur1: cur2;
   		while(node != null){
   			temp = new ListNode((node.val + carry) % 10);
   			carry = (node.val + carry) / 10;
   			cur.next = temp;
   			cur = cur.next;
   			node = node.next;
   		}
   		if(carry == 1){
   			cur.next = new ListNode(1);
   		}
   		return dummy.next;
   	}
   }

• Method 2: Optimization of Method 1.

   /**
    * 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 dummy = new ListNode(0);
   		ListNode cur = dummy;
   		int carry = 0;
   		while(l1 != null || l2 != null){
   			int sum = 0;
   			if(l1 != null && l2 != null){
   				sum = l1.val + l2.val + carry;
   				l1 = l1.next;
   				l2 = l2.next;
   			}else if(l1 != null){
   				sum = l1.val + carry;
   				l1 = l1.next;
   			}else{
   				sum = l2.val + carry;
   				l2 = l2.next;
   			}
   			cur.next = new ListNode(sum % 10);
   			carry = sum / 10;
   			cur = cur.next;
   		}
   		if(carry == 1) cur.next = new ListNode(1);
   		return dummy.next;
   	}
   }

Third Time

• Method 1: List

   /**
    * 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) {
           int carry = 0;
           ListNode dummy = new ListNode(0), cur = dummy;
           while(l1 != null || l2 != null){
               int sum = carry;
               if(l1 != null && l2 != null){
                   sum += l1.val + l2.val;
                   l1 = l1.next;
                   l2 = l2.next;
               }else if(l1 != null){
                   sum += l1.val;
                   l1 = l1.next;
               }else{
                   sum += l2.val;
                   l2 = l2.next;
               }
               carry = sum / 10;
               cur.next = new ListNode(sum % 10);
               cur = cur.next;
           }
           if(carry == 1)
               cur.next = new ListNode(1);
           return dummy.next;
       }
   }