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| 1 | +/*题目: |
| 2 | + |
| 3 | +链接:https://leetcode.com/problems/add-two-numbers/description/ |
| 4 | + |
| 5 | +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. |
| 6 | + |
| 7 | +You may assume the two numbers do not contain any leading zero, except the number 0 itself. |
| 8 | + |
| 9 | +Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) |
| 10 | +Output: 7 -> 0 -> 8 |
| 11 | + |
| 12 | +*/ |
| 13 | + |
| 14 | +/* |
| 15 | +解析: |
| 16 | +作者知乎链接:https://www.zhihu.com/people/bing-mo-43-95/activities |
| 17 | + |
| 18 | +解题思路:先把他当作两个普通的链表,进行相加,一个一个对应相加,很好实现。 |
| 19 | + 而在这题的难度是存在特殊的意义,即代表着两个数字的相加,这样的话就存在一个进位的问题。 |
| 20 | + 可以使用一个变量来存储进位,再加到下一位。 |
| 21 | + |
| 22 | +*/ |
| 23 | + |
| 24 | + |
| 25 | +/** |
| 26 | + * Definition for singly-linked list. |
| 27 | + * public class ListNode { |
| 28 | + * int val; |
| 29 | + * ListNode next; |
| 30 | + * ListNode(int x) { val = x; } |
| 31 | + * } |
| 32 | + */ |
| 33 | +class Solution { |
| 34 | + public ListNode addTwoNumbers(ListNode l1, ListNode l2) { |
| 35 | + ListNode result = new ListNode(0); |
| 36 | + ListNode head1,head2,head3;//两个临时变量 |
| 37 | + head1=l1; |
| 38 | + head2=l2; |
| 39 | + head3=result; |
| 40 | + //进位值digit |
| 41 | + int digit=0; |
| 42 | + while(head1!=null&&head2!=null) { |
| 43 | + ListNode LNode = new ListNode(head1.val+head2.val+digit); |
| 44 | + |
| 45 | + //如果相加结果大于等于10(即出现进位),则将val取10的余数。并将进位值设成1 |
| 46 | + if(LNode.val>=10){ |
| 47 | + LNode.val=LNode.val%10; |
| 48 | + digit = 1; |
| 49 | + } |
| 50 | + |
| 51 | + else{ |
| 52 | + //否则将数位归零 |
| 53 | + digit=0; |
| 54 | + } |
| 55 | + |
| 56 | + head3.next = LNode; |
| 57 | + head1=head1.next; |
| 58 | + head2=head2.next; |
| 59 | + head3=head3.next; |
| 60 | + |
| 61 | + |
| 62 | + |
| 63 | + |
| 64 | + } |
| 65 | + //解决两个数位不想等的情况 |
| 66 | + while(head1!=null){ |
| 67 | + ListNode LNode = new ListNode(head1.val+digit); |
| 68 | + if(LNode.val>=10){ |
| 69 | + LNode.val=LNode.val%10; |
| 70 | + digit = 1; |
| 71 | + } |
| 72 | + else{ |
| 73 | + digit=0;//将数位归零 |
| 74 | + } |
| 75 | + head3.next=LNode; |
| 76 | + head3=head3.next; |
| 77 | + head1=head1.next; |
| 78 | + |
| 79 | + } |
| 80 | + |
| 81 | + while(head2!=null){ |
| 82 | + ListNode LNode = new ListNode(head2.val+digit); |
| 83 | + if(LNode.val>=10){ |
| 84 | + LNode.val=LNode.val%10; |
| 85 | + digit = 1; |
| 86 | + } |
| 87 | + else{ |
| 88 | + digit=0;//将数位归零 |
| 89 | + } |
| 90 | + head3.next=LNode; |
| 91 | + head3=head3.next; |
| 92 | + head2=head2.next; |
| 93 | + } |
| 94 | + |
| 95 | + //最后一种情况,进位值仍然为1的时候 |
| 96 | + if(digit==1){ |
| 97 | + ListNode LNode = new ListNode(1); |
| 98 | + head3.next=LNode; |
| 99 | + } |
| 100 | + |
| 101 | + return result.next; |
| 102 | + } |
| 103 | +} |
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