给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。
请你将两个数相加,并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外,这两个数都不会以 0 开头。
输入:l1 = [2,4,3], l2 = [5,6,4] 输出:[7,0,8] 解释:342 + 465 = 807.
输入:l1 = [0], l2 = [0] 输出:[0]
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] 输出:[8,9,9,9,0,0,0,1]
l1
和l2
l3
,同时创建第三个指针指向l3
l1
和l2
,当前节点相加商
单独声明一个变量,用来下个节点相加余数
放到l3
当前节点l3
// 自己写的 var addTwoNumbers = function(l1, l2) { let l3 = new ListNode(0) let p1 = l1; let p2 = l2; let p3 = l3; let carry = 0; while(p1 || p2) { let val1 = p1? p1.val : 0; let val2 = p2 ? p2.val : 0; let total = val1 + val2 + carry; carry = Math.floor(total / 10); p3.val = total % 10; if((p1 && p1.next) || (p2 && p2.next)) { p3.next = new ListNode(0); p3 = p3.next; } if(p1) p1 = p1.next; if(p2) p2 = p2.next; } if(carry) { p3.next = new ListNode(carry) } return l3; };
// 视频中老师的解法 var addTwoNumbers = function(l1, l2) { let l3 = new ListNode(0) let p1 = l1; let p2 = l2; let p3 = l3; let carry = 0; while(p1 || p2) { let val1 = p1? p1.val : 0; let val2 = p2 ? p2.val : 0; let total = val1 + val2 + carry; carry = Math.floor(total / 10); p3.next = new ListNode(total % 10); if(p1) p1 = p1.next; if(p2) p2 = p2.next; p3 = p3.next; } if(carry) { p3.next = new ListNode(carry) } return l3.next; };
// leetcode的官方写法 var addTwoNumbers = function(l1, l2) { let head = null, tail = null; let carry = 0; while (l1 || l2) { const n1 = l1 ? l1.val : 0; const n2 = l2 ? l2.val : 0; const sum = n1 + n2 + carry; if (!head) { head = tail = new ListNode(sum % 10); } else { tail.next = new ListNode(sum % 10); tail = tail.next; } carry = Math.floor(sum / 10); if (l1) { l1 = l1.next; } if (l2) { l2 = l2.next; } } if (carry > 0) { tail.next = new ListNode(carry); } return head; };
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