93 lines
2.3 KiB
JavaScript
93 lines
2.3 KiB
JavaScript
/**
|
||
* 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}
|
||
*/
|
||
const addTwoNumbers = function (l1, l2) {
|
||
|
||
};
|
||
|
||
/*
|
||
直接遍历两个链表，列表的头节点代表的是各位，后一个节点代表的是十位，以此类推，所以只需定义一个
|
||
变量表示进位即可，把计算好的结果拼接成l3即可
|
||
*/
|
||
|
||
function ListNode(val, next) {
|
||
this.val = (val === undefined ? 0 : val);
|
||
this.next = (next === undefined ? null : next);
|
||
}
|
||
|
||
function f1(l1, l2) {
|
||
let carry = 0; // 进位
|
||
const dummyHead = new ListNode(0); // 哑节点，用于返回结果
|
||
let current = dummyHead; // 当前节点，用于构建求和之和的链表
|
||
|
||
// 遍历两个链表，直到，链表都为空，并且没有进位(只要有节点或者进位不等于0就需要创建新的求和节点)
|
||
|
||
while (l1 !== null || l2 !== null || carry !== 0) {
|
||
let sum = carry;
|
||
|
||
if (l1) {
|
||
sum += l1.val;
|
||
l1 = l1.next;
|
||
}
|
||
|
||
if (l2) {
|
||
sum += l2.val;
|
||
l2 = l2.next;
|
||
}
|
||
|
||
// 根据sum,计算出current节点的值和进位值
|
||
current.next = new ListNode(sum % 10);
|
||
current = current.next;
|
||
carry = Math.floor(sum / 10);
|
||
}
|
||
return dummyHead.next;
|
||
}
|
||
|
||
/*
|
||
将链表转换成字符串数字，之后把字符串数字相加，最后处理成一个新的列表
|
||
*/
|
||
function f2(l1, l2) {
|
||
let numStr1 = '';
|
||
let numStr2 = '';
|
||
|
||
// 将第一个链表反转并拼接为字符串
|
||
let cur1 = l1;
|
||
while (cur1 !== null) {
|
||
numStr1 = cur1.val + numStr1;
|
||
cur1 = cur1.next;
|
||
}
|
||
|
||
// 将第二个链表反转并拼接为字符串
|
||
let cur2 = l2;
|
||
while (cur2 !== null) {
|
||
numStr2 = cur2.val + numStr2;
|
||
cur2 = cur2.next;
|
||
}
|
||
|
||
// 将拼接的字符串转为数字并进行加法运算
|
||
const sum = BigInt(numStr1) + BigInt(numStr2); // 使用 BigInt 处理大数字
|
||
|
||
// 将求和结果转换为字符串
|
||
const sumStr = sum.toString();
|
||
|
||
// 将求和结果转换为链表
|
||
const dummyHead = new ListNode(0);
|
||
let current = dummyHead;
|
||
|
||
for (let i = sumStr.length - 1; i >= 0; i--) {
|
||
current.next = new ListNode(Number(sumStr[i]));
|
||
current = current.next;
|
||
}
|
||
|
||
return dummyHead.next; // 返回结果链表的头节点
|
||
}
|