0002. Add two numbers (M)

Add two numbers (M)

题目

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

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题意

以各位数倒序排列链表的形式给定两个整数，要求计算出两者之和，并同样以各位数倒序链表的形式返回结果。

思路

从两链表的个位数开始，依次计算各位数之和并判断记录进位。最后注意对最高位的进位进行处理。(注意先转换为两个整数再相加的方法会存在超过上限的问题)

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代码实现

Java

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);        // 创建头结点便于输出结果
        ListNode cur = head;
        int carry = 0;
        int a = 0, b = 0, c = 0;
        
        while (l1 != null || l2 != null) {
            a = (l1 == null)? 0 : l1.val;
            b = (l2 == null)? 0 : l2.val;
            c = a + b + carry;              
            carry = c / 10;             // 判断进位
            c = c % 10;
            cur.next = new ListNode(c);
            cur = cur.next;
            l1 = (l1 == null)? null : l1.next;
            l2 = (l2 == null)? null : l2.next;
        }
        // 最高位进位情况处理
        if (carry != 0) {
            cur.next = new ListNode(carry);
        }
        
        return head.next;
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
  let head = new ListNode()
  let p = head
  let carry = 0
  while (l1 !== null || l2 !== null) {
    let sum = carry
    sum += l1 === null ? 0 : l1.val
    sum += l2 === null ? 0 : l2.val
    p.next = new ListNode(sum % 10)
    carry = Math.floor(sum / 10)
    p = p.next
    if (l1 !== null) l1 = l1.next
    if (l2 !== null) l2 = l2.next
  }
  if (carry !== 0) {
    p.next = new ListNode(carry)
  }
  return head.next
}