leetcode : Add two numbers 解题报告
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Tag : dummy node
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 | /** * 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) { if (l1 == null && l2 == null ) { return null ; } if (l1 == null ) { return l2; } if (l2 == null ) { return l1; } ListNode head = new ListNode( 0 ); ListNode dummy = head; int carry = 0 ; while (l1 != null && l2 != null ) { int val = (l1.val + l2.val + carry) % 10 ; carry = (l1.val + l2.val + carry) / 10 ; ListNode node = new ListNode(val); head.next = node; head = node; l1 = l1.next; l2 = l2.next; } while (l1 != null ) { int val = (l1.val + carry) % 10 ; carry = (l1.val + carry) / 10 ; ListNode node = new ListNode(val); head.next = node; head = node; l1 = l1.next; } while (l2 != null ) { int val = (l2.val + carry) % 10 ; carry = (l2.val + carry) / 10 ; ListNode node = new ListNode(val); head.next = node; head = node; l2 = l2.next; } if (carry != 0 ) { ListNode node = new ListNode(carry); head.next = node; } return dummy.next; } } |
leetcode 最佳参考解法:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode c1 = l1; ListNode c2 = l2; ListNode sentinel = new ListNode( 0 ); ListNode d = sentinel; int sum = 0 ; while (c1 != null || c2 != null ) { sum /= 10 ; if (c1 != null ) { sum += c1.val; c1 = c1.next; } if (c2 != null ) { sum += c2.val; c2 = c2.next; } d.next = new ListNode(sum % 10 ); d = d.next; } if (sum / 10 == 1 ) d.next = new ListNode( 1 ); return sentinel.next; } } |
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