JS斩杀LeetCode（2）：Add Two Numbers

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.

示例：

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

原题链接：

https://leetcode.com/problems/add-two-numbers/#/description

简单讲一讲题目是什么意思。数字被逆序拆分保存在链表的各个节点中，现有两条链表，要将这两条链表相加返回新的链表。

以示例来说原数字为342和465，所以相加结果为807。

解题关键：① 怎么将每个节点连接组成一条新链表；② 考虑两节点相加时的进位。

解法①

/** * Definition for singly-linked list. * function ListNode(val) { *     this.val = val; *     this.next = null; * } *//** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */var addTwoNumbers = function(l1, l2) {    var result_list = null;    var cur_node = null;    var cur_carry = 0;    var v1 = 0;    var v2 = 0;        var checkSum = function(sum) {        if (parseInt(sum/10) > 0) {            cur_carry = parseInt(sum/10);            sum = sum % 10;        } else {            cur_carry = 0;        }        return sum;    };        var getValAndMoveToNext = function (list1, list2) {        v1 = list1 ? list1.val: 0;        v2 = list2 ? list2.val: 0;        l1 = list1 ? list1.next: null;        l2 = list2 ? list2.next: null;    };        if (l1 || l2) {        getValAndMoveToNext(l1, l2);        result_list = new ListNode(checkSum(v1+v2));        cur_node = result_list;        while (l1 || l2) {            getValAndMoveToNext(l1, l2);            var sum = v1 + v2 + cur_carry;            cur_node.next = new ListNode(checkSum(sum));            cur_node = cur_node.next;        }        if (cur_carry !== 0) {            cur_node.next = new ListNode(cur_carry);        }    }    return result_list;};

时间复杂度为O(n)，通过checkSum方法计算出当前节点的数值，以及进位情况：若sum除以10，商不为0，则需要考虑进位情况，当前节点的数值则除10取余。

result_list为最后返回的链表，cur_node的为当前所处的节点，通过以下代码将节点连成链表：

cur_node.next = new ListNode(checkSum(sum));cur_node = cur_node.next;

解法②

/** * Definition for singly-linked list. * function ListNode(val) { *     this.val = val; *     this.next = null; * } *//** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */var addOperation = function(l1, l2, carry) {    if (!l1 && !l2 && !carry) {        return null;    }    l1 = l1 || new ListNode(0);    l2 = l2 || new ListNode(0);    carry = carry || 0;    let l3 = new ListNode((l1.val + l2.val + carry)%10);    carry = parseInt((l1.val + l2.val + carry)/10);    if (l1.next || l2.next || carry) {        l3.next = addOperation(l1.next, l2.next, carry);        }    return l3;} var addTwoNumbers = function(l1, l2) {    return addOperation(l1, l2);};

递归的解法，时间复杂度为O(n)，和解法①的效率差不多，但代码更简洁一些。