Add Two Numbers (LeetCode)

https://leetcode.com/problems/add-two-numbers/

題目

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 contain a single digit. Add the two numbers and return it as a linked list.

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

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

答案

思路

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
function ListNode(val) {
    this.val = val;
    this.next = null;
}
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    var n1, n2, d = 0,
        v1, v2, o; //d進(jìn)位,o當(dāng)前位，n1 n1當(dāng)前節(jié)點(diǎn),v1 v2當(dāng)前節(jié)點(diǎn)的值
    n1 = l1;
    n2 = l2;
    v1 = n1 ? n1.val : 0;
    v2 = n2 ? n2.val : 0;
    o = (v1 + v2 + d) % 10; //當(dāng)前位
    d = Math.floor((v1 + v2 + d) / 10);
    var res = new ListNode(o);
    var node = res;
    while (n1 || n2 || d) {
        n1 = n1 ? n1.next : null;
        n2 = n2 ? n2.next : null;
        if (!n1 && !n2 && !d) break;
        v1 = n1 ? n1.val : 0;
        v2 = n2 ? n2.val : 0;
        o = (v1 + v2 + d) % 10; //當(dāng)前位
        d = Math.floor((v1 + v2 + d) / 10);
        node.next = new ListNode(o);
        node = node.next;
    }
    return res;
};

var l1 = new ListNode(2);
l1.next = new ListNode(4);
l1.next.next = new ListNode(3);
// l1.next.next.next = new ListNode(5);
var l2 = new ListNode(5);
l2.next = new ListNode(6);
l2.next.next = new ListNode(4);
console.log(addTwoNumbers(l1, l2))

核心方法進(jìn)階

var addTwoNumbers = function(l1, l2) {
    var n1, n2, d = 0,
        v1, v2, o; //d進(jìn)位,o當(dāng)前位，n1 n1當(dāng)前節(jié)點(diǎn),v1 v2當(dāng)前節(jié)點(diǎn)的值
    n1 = l1;
    n2 = l2;
    var res = new ListNode(0);
    var node=res;
    while (n1 || n2 || d) {
        v1 = n1 ? n1.val : 0;
        v2 = n2 ? n2.val : 0;
        o = (v1 + v2 + d) % 10; //當(dāng)前位
        d = Math.floor((v1 + v2 + d) / 10);
        node.next=new ListNode(o);
        node=node.next;
        n1 = n1?n1.next:null;
        n2 = n2?n2.next:null;
    }
    return res.next;
};

核心方法優(yōu)化

var addTwoNumbers = function(l1, l2) {
    //d進(jìn)位,o當(dāng)前位，l1 l1當(dāng)前節(jié)點(diǎn),v1 v2當(dāng)前節(jié)點(diǎn)的值
    var d = 0,
        v1, v2, o,
        res = new ListNode(0),
        node = res;
    while (l1 || l2 || d) {
        v1 = l1 ? l1.val : 0;
        v2 = l2 ? l2.val : 0;
        o = (v1 + v2 + d) % 10; //當(dāng)前位
        d = Math.floor((v1 + v2 + d) / 10);
        node.next = new ListNode(o);
        node = node.next;
        l1 = l1 ? l1.next : null;
        l2 = l2 ? l2.next : null;
    }
    return res.next;
};

繼續(xù)優(yōu)化

var addTwoNumbers = function(l1, l2) {
    var d = 0,//進(jìn)位
        res = {},
        node = res;
    while (l1 || l2 || d) {
        var v = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + d;//當(dāng)前值+進(jìn)位
        d = Math.floor(v / 10);
        node = node.next = new ListNode(v % 10);
        l1 = l1 ? l1.next : null;
        l2 = l2 ? l2.next : null;
    }
    return res.next;
};