标签：next length null lse val new === div return
这题思路跟2很相似。唯一的不同点在于，2只是正向做加法，但是445是在逆向做加法。这题我的思路是先用stack存住两个list的nodes，然后pop出来的时候做加法。这样就不需要操心reverse linked list这件事了。
1 /** 2 * @param {ListNode} l1 3 * @param {ListNode} l2 4 * @return {ListNode} 5 */ 6 var addTwoNumbers = function(l1, l2) { 7 let s1 = []; 8 let s2 = []; 9 while (l1 !== null) { 10 s1.push(l1.val); 11 l1 = l1.next; 12 } 13 while (l2 !== null) { 14 s2.push(l2.val); 15 l2 = l2.next; 16 } 17 18 let cur = new ListNode(0); 19 let sum = 0; 20 while (s1.length || s2.length) { 21 if (s1.length) { 22 sum += s1.pop(); 23 } 24 if (s2.length) { 25 sum += s2.pop(); 26 } 27 cur.val = sum % 10; 28 let head = new ListNode(Math.floor(sum / 10)); 29 head.next = cur; 30 cur = head; 31 sum = Math.floor(sum / 10); 32 } 33 if (cur.val === 0) { 34 return cur.next; 35 } else { 36 return cur; 37 } 38 };
[LeetCode] 445. Add Two Numbers II
标签：next length null lse val new === div return
原文地址：https://www.cnblogs.com/aaronliu1991/p/11666541.html
