题目
题目链接:https://leetcode-cn.com/problems/add-two-numbers-ii
给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。
你可以假设除了数字 0 之外,这两个数字都不会以零开头。
进阶:
如果输入链表不能修改该如何处理?换句话说,你不能对列表中的节点进行翻转。
示例:
输入:(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 8 -> 0 -> 7
解题记录
本来想偷个鸡想用数值和字符转换的方法求和,然后在写到链表:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public static ListNode addTwoNumbers(ListNode l1, ListNode l2){
StringBuilder s1 = new StringBuilder();
StringBuilder s2 = new StringBuilder();
while(l1!=null){
s1.append((char)(l1.val+48));
l1 = l1.next;
}
while(l2!=null){
s2.append((char)(l2.val+48));
l2 = l2.next;
}
char[] ss = (Long.parseLong(s1.toString())+Long.parseLong(s2.toString())+"").toCharArray();
ListNode res = null;
for (int i=ss.length-1; i>=0; --i){
ListNode node = new ListNode(ss[i]-48);
node.next = res;
res = node;
}
return res;
}
}
没想到太长会超出范围:
两眼一黑,只能老老实实出栈入栈了:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public static 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 res = null;
int carry = 0;
while (!s1.isEmpty() || !s2.isEmpty() || carry != 0){
int sum = (s1.isEmpty()?0:s1.pop()) + (s2.isEmpty()?0:s2.pop()) + carry;
carry = sum / 10;
ListNode node = new ListNode(sum%10);
node.next = res;
res = node;
}
return res;
}
}
来源:oschina
链接:https://my.oschina.net/u/4297014/blog/3235099