标签：lists nbsp mos return elf integer head while soft
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7
可看成是Add Two Numbers与Reverse Linked List的综合. 先reverse在逐个add, 最后把结果reverse回来.
Time Complexity: O(n). reverse O(n), add O(n).
Space: O(1). reverse O(1), add O(1).
AC Java:
/** * 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){ return l2; } if(l2 == null){ return l1; } l1 = reverseList(l1); l2 = reverseList(l2); ListNode cur = l1; int len1 = 0; while(cur != null){ len1++; cur = cur.next; } cur = l2; int len2 = 0; while(cur != null){ len2++; cur = cur.next; } ListNode dummy = new ListNode(0); if(len1 > len2){ dummy.next = l1; }else{ dummy.next = l2; } cur = dummy; int carry = 0; while(l1 != null || l2 != null){ if(l1 != null){ carry += l1.val; l1 = l1.next; } if(l2 != null){ carry += l2.val; l2 = l2.next; } cur.next.val = carry%10; cur = cur.next; carry /= 10; } if(carry != 0){ cur.next = new ListNode(1); } ListNode nxt = dummy.next; ListNode newHead = reverseList(nxt); nxt.next = null; return newHead; } private ListNode reverseList(ListNode head){ if(head == null || head.next == null){ return head; } ListNode tail = head; ListNode cur = head; ListNode pre; ListNode temp; while(tail.next != null){ pre = cur; cur = tail.next; temp = cur.next; cur.next = pre; tail.next = temp; } return cur; } }
LeetCode-Microsoft-Add Two Numbers II
标签：lists nbsp mos return elf integer head while soft
原文地址：https://www.cnblogs.com/incrediblechangshuo/p/8972208.html
