445. Add Two Numbers II 链表中的数字求和
[抄题]:
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
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
最后看第一位是不是0,不是0才能返回。
[思维问题]:
顺序是反的,不知道用stack
通过sum / 10的方法可以把十位取出来,不用清空,每次除10一直加就行了
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
顺序是反的就要用stack求和
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 先初始化一个待添加列表list,然后余数都往后面加。求和的头都放在head中。
[二刷]:
- list.val = sum % 10; 可以直接往用等号list后面加点,不知道怎么来的
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
通过sum / 10的方法可以把十位取出来,不用清空,每次除10一直加就行了。倒序相加用stack
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
/**
* 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) {
//initialization: 2 stacks
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>(); //corner case
if (l1 == null && l2 == null) return null; //put into stacks
while (l1 != null) {
stack1.add(l1.val);
l1 = l1.next;
} while (l2 != null) {
stack2.add(l2.val);
l2 = l2.next;
} //while loop
//get sum, val, head, append the previous val to head, give head to val, reset sum
ListNode list = new ListNode(0);
int sum = 0;
while (!stack1.isEmpty() || !stack2.isEmpty()) {
//get sum
if (!stack1.isEmpty()) sum += stack1.pop();
if (!stack2.isEmpty()) sum += stack2.pop(); //get the val node
list.val = sum % 10; ListNode head = new ListNode(sum / 10);
//append the previous val to head, head is ready
head.next = list;
//give head to val
list = head; sum /= 10;
} //return
return list.val == 0 ? list.next : list;
}
}
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