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 contains a single digit. Add the two numbers and return the sum as a linked list.

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

example

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Input: l1 = [0], l2 = [0]
Output: [0]

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

我的思路

//leetcode submit region begin(Prohibit modification and deletion)

// class ListNode {
//     val: number
//     next: ListNode | null
//     constructor(val?: number, next?: ListNode | null) {
//         this.val = (val === undefined ? 0 : val)
//         this.next = (next === undefined ? null : next)
//     }
// }

// function buildBinaryTree(arr: Array<number>): ListNode|null {
//     let head = new ListNode(arr[0]);
//     let cur = head;
//     arr.slice(1).forEach(element => {
//         cur.next = new ListNode(element);
//         cur = cur.next;
//     });
//     return head;
// }


function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
    // 扔两个指针
    let ptr1: ListNode | null = l1,
        ptr2: ListNode | null = l2;
    // head ptr
    // 弄个head出来 方便返回 而且循环里可以少写几行代码(不用判空)
    let head = new ListNode(), ptr = head;
    // 进位
    let t = 0;
    // 全部为null时stop
    while (ptr1 || ptr2) {
        ptr.next = new ListNode();
        ptr = ptr.next;
        // 当前位总和
        let sum = ((ptr1 ? ptr1.val : 0) + (ptr2 ? ptr2.val : 0)) + t;
        // 当前位
        let cur = sum % 10;
        // 存储进位
        t = Number.parseInt(String(sum / 10));
        // 存储当前位
        ptr.val = cur;
        ptr1 = ptr1 && ptr1.next;
        ptr2 = ptr2 && ptr2.next;
    }
    // 如果t最后是1 需要再创建一个节点加上去
    if (t) {
        ptr.next = new ListNode(t);
    }
    return head.next;
};
// 测试代码
// console.log(addTwoNumbers(buildBinaryTree([2,4,3]), buildBinaryTree([5,6,4])));
// console.log(addTwoNumbers(buildBinaryTree([9,9,9,9,9,9,9]), buildBinaryTree([9,9,9,9])))
// console.log(addTwoNumbers(buildBinaryTree([0]), buildBinaryTree([0])))

最优解

最优解应该是我这个了，但是我平时绝对不会写这种可读性这么垃圾的代码

肯定是new一个数组，在数组里加完再转成链表

另外这个思路也不错，起码短

/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    let s1 = JSON.stringify(l1).match(/\d/g).reverse().join(''),
        s2 = JSON.stringify(l2).match(/\d/g).reverse().join('')

    sum = BigInt(s1)+BigInt(s2)

    return [...sum.toString()].reduce((acc,v)=>({val: v, next: acc}), null)
};