【Leetcode】002 Add Two Numbers

【Leetcode】002 Add Two Numbers

题目

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

思路

使用dummyHead避免写重复的代码，非常巧妙

实现

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers0(ListNode l1, ListNode l2) {
int val = (l1.val + l2.val) % 10;
int increment = (l1.val + l2.val) / 10;
ListNode head = new ListNode(val);

ListNode currNode = head;
l1 = l1.next;
l2 = l2.next;
for (ListNode n1 = l1, n2 = l2; l1 != null && l2 != null; l1 = l1.next, l2 = l2.next){
int result = l1.val + l2.val + increment;
currNode.next = new ListNode(result % 10);
increment = result / 10;
currNode = currNode.next;
}

for (; l1 != null; l1 = l1.next){
int result = l1.val + increment;
currNode.next = new ListNode(result % 10);
increment = result / 10;
currNode = currNode.next;
}

for (; l2 != null; l2 = l2.next){
int result = l2.val + increment;
currNode.next = new ListNode(result % 10);
increment = result / 10;
currNode = currNode.next;
}

if (increment == 1){
currNode.next = new ListNode(1);
}
return head;
}

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);  // 第二个结点是链表的头结点
int increment = 0;
ListNode currNode = dummyHead;

for (ListNode n1 = l1, n2 = l2; l1 != null || l2 != null;){
int x = l1 != null ? l1.val : 0;
int y = l2 != null ? l2.val : 0;

int result = x + y + increment;
currNode.next = new ListNode(result % 10);
increment = result / 10;
currNode = currNode.next;

if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}

if (increment == 1){
currNode.next = new ListNode(1);
}

return dummyHead.next;
}
}