Problem 2: Add Two Numbers
思路
这道题就是“逐位加法”的例子:进位 carrier,每个位置顺序往上加,超过 10 就进位。
carrier 最后如果大于 0 的话,扩大原来的 size,把这一位补上 1
/**
* 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 && l2 == null) {
return null;
}
ListNode dummy = new ListNode(0);
ListNode head = dummy;
int carrier = 0;
while (l1 != null && l2 != null) {
int sum = carrier + l1.val + l2.val;
head.next = new ListNode(sum % 10);
carrier = sum / 10;
l1 = l1.next;
l2 = l2.next;
head = head.next;
}
while (l1 != null) {
int sum = carrier + l1.val;
head.next = new ListNode(sum % 10);
carrier = sum / 10;
l1 = l1.next;
head = head.next;
}
while (l2 != null) {
int sum = carrier + l2.val;
head.next = new ListNode(sum % 10);
carrier = sum / 10;
l2 = l2.next;
head = head.next;
}
if (carrier != 0) {
head.next = new ListNode(carrier);
}
return dummy.next;
}
}
易错点
记得设置 dummy node,否则会对整个 list 失去“控制”
第一次 l1, l2 循环完了以后,是用 while 接着看剩下的
while (l1 != null) { int sum = carrier + l1.val; head.next = new ListNode(sum % 10); carrier = sum / 10; l1 = l1.next; head = head.next; }
我当时用的是 if 一直没找到错误。其实在 linked list 的题目里面,最后都是用 while 来扫除余下的
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