链表两数想加问题优化点

初始版本


/*
 * @lc app=leetcode.cn id=2 lang=javascript
 *
 * [2] 两数相加
 */

// @lc code=start
/**
 * 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) {
  const head = new ListNode(0, null);
  let p = head;

  let up = 0;
  while (l1 && l2) {
    const sum = l1.val + l2.val + up;

    const cur = sum % 10;

    const tmp = new ListNode(cur, null);

    up = Math.floor(sum / 10);

    p.next = tmp;
    p = tmp
    l1 = l1.next;

    l2 = l2.next
  }

  let rest = l1 || l2;

  while(rest){
    const sum = rest.val + up;

    const cur = sum % 10;

    const tmp = new ListNode(cur, null);

    up = Math.floor(sum / 10);

    p.next = tmp;
    p = tmp

    rest = rest.next;
  }

  if(up){
    const tmp = new ListNode(up, null);

    p.next = tmp;
  }

  return head.next;
};
// @lc code=end

这个版本虽然也能实现，但是代码量较多，其实有一些重复的逻辑

优化版本


/*
 * @lc app=leetcode.cn id=2 lang=javascript
 *
 * [2] 两数相加
 */

// @lc code=start
/**
 * 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) {
  const head = new ListNode(0, null);
  let p = head;

  let up = 0;
  while (l1 || l2 || up) {
    let sum = 0;

    if (l1) {
      sum += l1.val;
      l1 = l1.next;
    }

    if (l2) {
      sum += l2.val;
      l2 = l2.next;
    }

    sum += up;

    const cur = sum % 10;

    const tmp = new ListNode(cur, null);

    up = Math.floor(sum / 10);

    p.next = tmp;
    p = tmp
  }
  return head.next;
};
// @lc code=end

将l1,l2,up的逻辑在一个循环里处理，因为其实际逻辑都是一样的，创建新的节点并链接到旧节点