题目
给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。你可以假设除了数字 0 之外,这两个数都不会以 0 开头。请你将两个数相加,并以相同形式返回一个表示和的链表。
第一种:直接链表加
public class AddTwoNumbers {
public class ListNode {
int val;
ListNode next;
ListNode() {}
ListNode(int val) { this.val = val; }
ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int tem = 0; //进位值
int val = l1.val + l2.val; //求同位置之和
tem = val/10; //求进位
ListNode head = new ListNode(val % 10);
ListNode ll1 = l1.next;
ListNode ll2 = l2.next;
ListNode head1 = head;
while (ll1!=null && ll2!=null) {
int val1 = ll1.val + ll2.val + tem; //加上进位值
head1.next = new ListNode(val1 % 10); //求余
tem = val1/10;
ll1 = ll1.next;
ll2 = ll2.next;
head1 = head1.next;
}
ListNode l = ll1!=null ? ll1 : ll2; //剩下的继续计算
while (l!=null) {
int val1 = l.val + tem;
head1.next = new ListNode(val1 % 10);
tem = val1/10;
l = l.next;
head1 = head1.next;
}
if (tem!=0) { //可能有进位
head1.next = new ListNode(tem);
}
return head;
}
public static void main(String[] args) {
AddTwoNumbers addTwoNumbers = new AddTwoNumbers();
ListNode node1 = addTwoNumbers.new ListNode(1);
node1.next = addTwoNumbers.new ListNode(2);
node1.next.next = addTwoNumbers.new ListNode(3);
ListNode node2 = addTwoNumbers.new ListNode(7);
node2.next = addTwoNumbers.new ListNode(8);
node2.next.next = addTwoNumbers.new ListNode(9);
//1,2,3
//7,8,9
//321+987=1308
//8,0,3,1
ListNode node = addTwoNumbers.addTwoNumbers(node1, node2);
}
}