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一、题目

二、思路

• 加法运算是从低位开始，向高位进位，因此需要将两个链表进行反转，再进行对齐后的相加操作。
• 力扣 2. 两数相加

三、题解

/*** Definition for singly-linked list.* 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; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode ans;l1 = reverseList(l1);l2 = reverseList(l2);return reverseList(addTwo(l1, l2, 0));}// 反转链表public ListNode reverseList(ListNode head) {ListNode pre = null, cur = head;while (cur != null) {ListNode tmp = cur.next;cur.next = pre;pre = cur;cur = tmp;}return pre;}// 求解两数之和 public ListNode addTwo(ListNode l1, ListNode l2, int carry) {// 递归出口if (l1 == null && l2 == null) {return carry != 0 ? new ListNode(carry) : null;}// 确保 l1 是较长的if (l1 == null) {l1 = l2;l2 = null;}int sum = l1.val + (l2 != null ? l2.val : 0) + carry;l1.val = sum % 10;l1.next = addTwo(l1.next, l2 != null ? l2.next : null, sum / 10);return l1;}
}