Leetcode 2. Add Two Numbers
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题目链接
https://leetcode.com/problems/add-two-numbers/description/
题目描述
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order 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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
题解
依次遍历两个链表,计算对应位的和,如果有进位,计算进位。
XBB Time:刚开始写的check函数是把当前节点的前一个节点(记为pre),带进函数里面的,但是总是过不了,感觉是复制了一份传入,函数内的修改了值,但是外层函数中的pre并没有变,这就又回到了到底是值传递还是引用传递,其实是把引用复制了一份,然后传递进去,操作的是同一个内存地址,对该内存地址的修改是可见的,但是函数里的pre已经指向其他的地址了,外层函数的pre是不会变的。
代码
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
int count = 0;
ListNode start = new ListNode(0);
ListNode pre = start;
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
while (l1 != null && l2 != null) {
int val = l1.val + l2.val + count;
count = val / 10;
ListNode node = new ListNode(val % 10);
pre.next = node;
pre = node;
l1 = l1.next;
l2 = l2.next;
}
// check(l1, pre);
// check(l2, pre);
check(l1);
check(l2);
if (count != 0) {
ListNode node = new ListNode(count);
pre.next = node;
pre = node;
}
return start.next;
}
/**
* 开始把count带入了函数,有的用例过不了,count是值传递
**/
public void check(ListNode l) {
while (l != null) {
if (count == 0) {
pre.next = l;
break;
}
int val = l.val + count;
count = val / 10;
ListNode node = new ListNode(val % 10);
pre.next = node;
pre = node;
l = l.next;
}
}
}
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