445.两数相加-ii.md

445.两数相加-ii

/*
 * @lc app=leetcode.cn id=445 lang=typescript
 *
 * [445] 两数相加 II
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * 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 addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {}

// @lc code=end

解法 1: 模拟

function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
  const arr1: ListNode[] = [],
    arr2: ListNode[] = []
  while (l1) {
    arr1.push(l1)
    l1 = l1.next
  }
  while (l2) {
    arr2.push(l2)
    l2 = l2.next
  }
  const arr3: ListNode[] = []
  let pre = 0
  while (arr1.length || arr2.length || pre) {
    const num = (arr1.pop()?.val ?? 0) + (arr2.pop()?.val ?? 0) + pre
    const cur = new ListNode(num % 10)
    arr3.push(cur)
    pre = Math.floor(num / 10)
  }
  arr3.reverse()
  let root: ListNode | null = null,
    node = root
  for (let i = 0; i < arr3.length; i++) {
    if (!node) node = root = arr3[i]
    else {
      node.next = arr3[i]
      node = arr3[i]
    }
  }
  return root
}

Case

test.each([
  { input: { l1: [7, 2, 4, 3], l2: [5, 6, 4] }, output: [7, 8, 0, 7] },
  { input: { l1: [2, 4, 3], l2: [5, 6, 4] }, output: [8, 0, 7] },
  { input: { l1: [0], l2: [0] }, output: [0] },
])('input: l1 = $input.l1, l2 = $input.l2', ({ input: { l1, l2 }, output }) => {
  expect(addTwoNumbers(LinkedList.deserialize(l1), LinkedList.deserialize(l2))).toEqual(LinkedList.deserialize(output))
})