leetcode 2 Add Two Numbers

题目内容

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.

分析过程

题目归类：
链表，迭代
题目分析：
最好的情况是能够走一趟，想到使用递归可以层层深入，先计算最外的数据。然后向内传进位来计算。麻烦的是如何判断两个链的长度
注意链之间和进位之间的关系。
边界分析：
- 空值分析
  当都为null是返回null
  当一方为null返回另一方。
- 循环边界分析
方法分析：
- 数据结构分析
  链表需要控制什么时候指针指向null
  递归需要考虑返回的内容（当指向null时返回null，当一方为null返回另一方）
- 状态机
- 状态转移方程
- 最优解
测试用例构建
[0],[0];
null,null;

代码实现

/**
 * 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) {
        return addPlus(l1,l2,0);
    }
    public ListNode addPlus(ListNode l1, ListNode l2, int carry){
        if(l1==null && l2 ==null&&carry!=0){
            return new ListNode(carry);
        }else if(l1 == null && l2 == null&& carry == 0)
            return null;
        else if(l1 == null&&carry==0)
            return l2;
        else if(l1 == null&&carry!=0){
            int sum = l2.val +carry;
            l2.val = sum%10;
            l2.next = addPlus(l1,l2.next,sum/10);
            return l2;
         }else if (l2 == null&&carry==0)
            return l1;
         else if(l2 == null&&carry!=0){
            int sum = l1.val +carry;
            l1.val = sum%10;
            l1.next=addPlus(l1.next,l2,sum/10);
            return l1;
         }
        int sum = l1.val+l2.val+carry;
        
        ListNode head = new ListNode(sum%10);
        head.next = addPlus(l1.next,l2.next,sum/10);
        return head;
        
    }
}

效率提高

使用非递归的方式可以获得更加简洁的代码，并且不需要很多if/else

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

拓展问题

Multiply Strings
Add Binary
Sum of Two Integers
Add Strings
Add Two Numbers II
Add to Array-Form of Integer