[Leetcode]2. 两数相加

题目描述

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。

请你将两个数相加,并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外,这两个数都不会以 0 开头。

  • 示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.
  • 示例 2:
输入:l1 = [0], l2 = [0]
输出:[0]
  • 示例 3:
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]

本题考查链表相关知识。

java解法

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;
  }

  static ListNode of(int[] nums) {
    ListNode dummyHead = new ListNode(-1);
    for (int i = nums.length - 1; i >= 0; i--) {
      dummyHead.next = new ListNode(nums[i], dummyHead.next);
    }
    return dummyHead.next;
  }

  @Override
  public String toString() {
    ListNode cur = this;
    StringBuilder buffer = new StringBuilder();
    while (cur != null) {
      buffer.append(cur.val).append("->");
      cur = cur.next;
    }
    return buffer.toString();
  }
}
class Solution {

  public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode curL1 = l1;
    ListNode curL2 = l2;
    //进位
    int carry = 0;
    ListNode dummyHead = new ListNode(-1);
    ListNode cur = dummyHead;
    while (curL1 != null || curL2 != null) {
      int val1 = curL1 == null ? 0 : curL1.val;
      int val2 = curL2 == null ? 0 : curL2.val;
      int val = val1 + val2 + carry;
      cur.next = new ListNode(val % 10);
      carry = val / 10;
      if (curL1 != null) {
        curL1 = curL1.next;
      }
      if (curL2 != null) {
        curL2 = curL2.next;
      }
      cur = cur.next;
    }
    if (carry == 1) {
      cur.next = new ListNode(1);
    }
    return dummyHead.next;
  }


  public static void main(String[] args) {
    ListNode l1 = ListNode.of(new int[]{9, 9, 9, 9, 9, 9, 9});
    ListNode l2 = ListNode.of(new int[]{9, 9, 9, 9});
    ListNode listNode = new Solution().addTwoNumbers(l1, l2);
    System.out.println(l1);
    System.out.println(l2);
    System.out.println(listNode);
  }
}
posted @ 2021-01-12 20:28  strongmore  阅读(62)  评论(0编辑  收藏  举报