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Copy pathAdd Two Numbers.java
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executable file
·119 lines (103 loc) · 2.75 KB
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M
1519711343
tags: Linked List, Math
LinkedList都已经反转好了，直接做.
遍历两个l1,l2把carry-on处理好，每次生成一个新node，最后检查carry-on.
跟Add Binary的理解方式一模一样.
注意:
Linked List 没有天然size.
用DummyNode(-1).next来hold住结果.
```
/*
You have two numbers represented by a linked list,
where each node contains a single digit.
The digits are stored in reverse order,
such that the 1's digit is at the head of the list.
Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.
Tags Expand
Cracking The Coding Interview Linked List High Precision
*/
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
/*
Thoughts:
The list has been reversed, just add them up.
Append one more ListNode if there is an carry.
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
int carry = 0;
ListNode node = new ListNode(-1);
ListNode head = node;
// Add l1 and l2
while (l1 != null || l2 != null) {
int sum = carry;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
carry = sum / 10;
sum = sum % 10;
node.next = new ListNode(sum);
node = node.next;
}
// Finish adding carry
if (carry != 0) {
node.next = new ListNode(carry);
}
return head.next;
}
}
// Previous solution
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
ListNode rst = new ListNode(0);
ListNode dummy = rst;
int carrier = 0;
//while
while (l1 != null || l2 != null) {
if (l1 != null) {
carrier += l1.val;
l1 = l1.next;
}
if (l2 != null) {
carrier += l2.val;
l2 = l2.next;
}
rst.next = new ListNode(carrier % 10);
carrier = carrier / 10;
rst = rst.next;
}
//check the carrier
if (carrier == 1) {
rst.next = new ListNode(1);
}
return dummy.next;
}
}
```