image.png

题解

  1. /**
  2.  * Definition for singly-linked list.
  3.  * class ListNode {
  4.  *     val: number
  5.  *     next: ListNode | null
  6.  *     constructor(val?: number, next?: ListNode | null) {
  7.  *         this.val = (val===undefined ? 0 : val)
  8.  *         this.next = (next===undefined ? null : next)
  9.  *     }
  10.  * }
  11.  */
  13. function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
  14.     // 头节点             当前节点
  15.     let head = null; let target = null;
  16.       // 累加值  传递到下一次循环
  17.     let carry = 0;
  18.     while (l1 || l2) {
  19.         let val1 = l1 && l1.val || 0;
  20.         let val2 = l2 && l2.val || 0;
  21.         let sum = val1 + val2 + carry;
  22.           // 判断是否第一次
  23.         if (!head) {
  24.             head = target = new ListNode(sum % 10);
  25.         } else {
  26.             target.next = new ListNode(sum % 10);
  27.             target = target.next;
  28.         }
  29.           // 重置累加值
  30.         carry = Math.floor(sum / 10);
  31.         if (l1) {
  32.             l1 = l1.next;
  33.         }
  34.         if (l2) {
  35.             l2 = l2.next;
  36.         }
  37.     }
  38.       // 此时如果carry > 0 说明要加一位 也就是新加一个节点
  39.     if (carry > 0) {
  40.         target.next = new ListNode(carry);
  41.     }
  42.     return head;
  43. };

相似题目

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思路

解法与上题相似，不过反过来了， 我们可以通过两个栈（stack）来记录遍历两条链表的值，
通过一个max变量 比较两个栈的大小 取最大值进行遍历，从尾部构建出整个链表。

代码

/**
 * 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 {
    let target = null;
    let a = null;
    let stack1 = [], stack2 = [];
    while (l1) {
        stack1.push(l1.val);
        l1 = l1.next;
    }
    while (l2) {
        stack2.push(l2.val);
        l2 = l2.next;
    }

    let carry = 0;
    let arr = [];
    let sum = 0;
    let max = Math.max(stack1.length, stack2.length);
    for (let i = 0; i < max; i++) {
        let val1 = stack1.pop() || 0;
        let val2 = stack2.pop() || 0;
        sum = val1 + val2 + carry;
        a = new ListNode(sum % 10);
        a.next = target;
        target = a;
        carry = Math.floor(sum / 10);
    }
    let x = null;
    if (carry > 0) {
        x = new ListNode(carry);
        x.next = target;
        console.log(x.val);
    }
    console.log(target.val);
    return target ? carry ? x : target : null;
};