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
Output: 7 -> 0 -> 8
A:
Mistakes:
1: 忘记 给l1, l2 也增加runner, --------- 这里其实可以直接用l1,l2当做runner,毕竟java是传引用的。
2: 应该是在计算完毕 sun之后,再update carry的值。
3: 当carry 最后 进一位的时候, 我们copy 其余的产品的时候, 忘了把其也加上。 哎~~~~~
这道题,就是考的细心啊~~~~~~~~~
4: 为什么,( 3+5 )%10 的结果总是1 呢? 搞得carry 变成1 了。 哎 ---------- 计算carry的时候,把r2 写成r1了。
5: 当carry 最后加完了, 还是1 的时候,我们要再生成一个节点。 ╮(╯▽╰)╭, Tiger,你这样不行啊~~~~~~~~~~~~SB错误太多了
--------------------------3 rd pass--------------use only one header, for newly created list header 跟第一遍的写法类似---------------------- 好消息是, 只有几个拼写错误, 不好的是, 还是不能bug free--------------------
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode header = new ListNode(0); ListNode tail = header; int carry = 0; while(carry !=0 || l1!=null || l2!=null){ int a =0,b=0,c =0; if( l1!=null ){ a = l1.val; l1 = l1.next; } if( l2!=null ){ b = l2.val; l2 = l2.next; } ListNode tmp = new ListNode((a+b+carry)%10); tail.next = tmp; tail = tail.next; carry = (a+b+carry) /10; } return header.next; } }
-------------------4 th pass---------------recursive way---------------
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { return helper(l1,l2,0); } private ListNode helper(ListNode l1,ListNode l2, int carry){ ListNode curNode; if(l1==null && l2 == null){ return carry == 0?null: new ListNode(carry); }else if(l1 == null){ curNode = new ListNode((carry+l2.val)%10); carry = (carry+l2.val)/10; curNode.next = helper(null,l2.next,carry); }else if(l2 == null){ curNode = new ListNode((carry+l1.val)%10); carry = (carry+l1.val)/10; curNode.next = helper(l1.next,null,carry); }else{ curNode = new ListNode((carry+l1.val+l2.val)%10); carry = (carry+l1.val+l2.val)/10; curNode.next = helper(l1.next,l2.next,carry); } return curNode; } }
----------------------递归调用的 简化版本------------
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { return helper(l1,l2,0); } private ListNode helper(ListNode l1,ListNode l2, int carry){ if(l1==null && l2 == null) return carry == 0?null: new ListNode(carry); int sum = (l1==null?0:l1.val) + (l2==null?0:l2.val)+carry; ListNode curNode = new ListNode((sum)%10); curNode.next = helper(l1==null?null:l1.next,l2==null?null:l2.next,sum/10); return curNode; } }
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