LeetCode445. Add Two Numbers II

You are given two linked lists representing two non-negative numbers. 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

这道题就是让做两个链表数据结构的大整数加法。

没啥好说的，限制条件里面说不允许对链表进行reverse，但是这边没有说必须要O（1）的空间复杂度，因此直接用一个Stack，完成从后往前的遍历。

代码贴上来：

/** * 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;        int length1 = getLength(l1);        int length2 = getLength(l2);        if(length1 < length2){            ListNode temp = l2;            l2 = l1;            l1 = temp;            int tmp = length2;            length2 = length1;            length1 = tmp;        }        ListNode p = l1;        for(int i = 0;i<length1-length2;i++){            p = p.next;        }        for(int i = 0;i<length2;i++){            p.val+=l2.val;            p = p.next;            l2 = l2.next;        }        p = l1;        Stack<ListNode> stack = new Stack<>();        while(p!=null){            stack.push(p);            p = p.next;        }        while(!stack.isEmpty()){            ListNode n = stack.pop();            if(stack.isEmpty())                break;            if(n.val>=10){                n.val%=10;                stack.peek().val+=1;            }        }        if(l1.val>=10){            l1.val%=10;            ListNode n = new ListNode(1);            n.next = l1;            l1 = n;        }        return l1;    }        private int getLength(ListNode l){        int counter = 0;        while(l!=null){            counter++;            l = l.next;        }        return counter;    }}

代码感觉写的还是有些乱，有写地方是一边改一边写的，没有做好优化，但是时间也不是很充足，将就这样吧。