标签：cannot || 元素 style other time for div exce
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
M1: 先把两个链表反转，相加之后再反转
time: O(n), space: O(n)
/** * 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 dummy = new ListNode(0); ListNode prehead = dummy; int remainder = 0; ListNode p1 = reverse(l1), p2 = reverse(l2); while(p1 != null || p2 != null) { int sum = remainder; if(p1 != null) { sum += p1.val; p1 = p1.next; } if(p2 != null) { sum += p2.val; p2 = p2.next; } dummy.next = new ListNode(sum % 10); remainder = sum / 10; dummy = dummy.next; } if(remainder != 0) { dummy.next = new ListNode(remainder); } ListNode res = reverse(prehead.next); return res; } private ListNode reverse(ListNode head) { ListNode prev = null, cur = head; while(cur != null) { ListNode nextnode = cur.next; cur.next = prev; prev = cur; cur = nextnode; } return prev; } }
M2: follow-up 不能反转链表
用两个stack分别存储链表元素，再相加。
注意如果相加的话，得到的结果是反的，所以在相加的时候，就要边反转此结果链表
detail: 把dummy的值设为sum%10，tmp是新节点，其值为carry，tmp指向dummy，然后dummy向前移动一位，即dummy = tmp
最后还要判断一下leading number是不是0，如果为0，则返回下一个节点
time: O(n), space: O(n)
/** * 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) { Stack<Integer> s1 = new Stack<>(); Stack<Integer> s2 = new Stack<>(); while(l1 != null) { s1.push(l1.val); l1 = l1.next; } while(l2 != null) { s2.push(l2.val); l2 = l2.next; } ListNode dummy = new ListNode(0); int carry = 0; while(!s1.isEmpty() || !s2.isEmpty()) { int sum = carry; if(!s1.isEmpty()) { sum += s1.pop(); } if(!s2.isEmpty()) { sum += s2.pop(); } dummy.val = sum % 10; carry = sum / 10; ListNode tmp = new ListNode(carry); tmp.next = dummy; dummy = tmp; } return dummy.val == 0 ? dummy.next : dummy; } }
445. Add Two Numbers II - Medium
标签：cannot || 元素 style other time for div exce
原文地址：https://www.cnblogs.com/fatttcat/p/10122443.html
