新手做LeetCode 2 Add Two Numbers
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/**
*You are given two linked lists representing two non-negative numbers.
*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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
*
*/
public class N2_AddTwoNumbers {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode firstNode = new ListNode(0);
if(l1 == null && l2 == null)
{
return new ListNode(0);
}
if(l1 == null)
{
firstNode.val = l2.val;
firstNode.next = this.add(null, l2.next, firstNode, 0);
}
if(l2 == null){
firstNode.val = l1.val;
firstNode.next = this.add(l1.next, null, firstNode, 0);
}
if(l1.val + l2.val >= 10)
{
firstNode.val = this.add(l1.val, l2.val);
firstNode.next = this.add(l1.next, l2.next, firstNode, 1);
}
else
{
firstNode.val = this.add(l1.val, l2.val);
firstNode.next = this.add(l1.next, l2.next, firstNode, 0);
}
return firstNode;
}
private ListNode add(ListNode l1, ListNode l2,ListNode prevNode,int carry)
{
if(l1 == null && l2 == null && carry == 0)
{
return null;
}
else if( l1 == null && l2 == null && carry == 1)
{
prevNode.next = new ListNode(1);
return null;
}
if(l1 == null)
{
prevNode.next = new ListNode(this.add(0, l2.val + carry));
if(l2.val + carry >= 10)
{
carry = 1;
}
else
{
carry = 0;
}
return this.add(null, l2.next, prevNode.next,carry);
}
if(l2 == null)
{
prevNode.next = new ListNode(this.add(l1.val + carry, 0));
if(l1.val + carry >= 10)
{
carry = 1;
}
else
{
carry = 0;
}
return this.add(l1.next, null, prevNode.next,carry);
}
if(l1.val + l2.val + carry >= 10)
{
prevNode.next = new ListNode(this.add(l1.val, l2.val + carry));
return this.add(l1.next , l2.next, prevNode.next,1);
}
else
{
prevNode.next = new ListNode(this.add(l1.val, l2.val + carry));
return this.add(l1.next, l2.next, prevNode.next,0);
}
}
private int add(int a,int b)
{
return (a + b) % 10;
}
public static void main(String[] args) {
}
private class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
}
给了两个链表,每个节点的值都是一个一位数,要求将这两个链表相加,得到的链表要求是一位数,要进位。考虑不周到的地方就在于进位。
在计算的时候都必须带上进位,这样就可以了。
0 0
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