两数相加

[louzhedong]

习题

出处：LeetCode 算法第2题

给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

思路

简单的链表，每相加两个链表的值，将进位放存放在carry里，余数存放在remainder里。进行下一次计算时，需要加上carry的值。另一个易错点就是当其中一个链表已经到尾部，而carry的值为1时的情况。

解答

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
  var result = new ListNode(0);
  var head = result;
  var carry = 0;
  var remainder = 0;
  while (l1 != null && l2 != null) {
    var temp = l1.val + l2.val + (carry ? 1 : 0);
    if (temp >= 10) {
      carry = 1;
      remainder = temp % 10;
    } else {
      carry = 0;
      remainder = temp;
    }
    var add = new ListNode(remainder);
    result.next = add;
    result = result.next;
    l1 = l1.next;
    l2 = l2.next;
  }
  if (l1 == null && l2 == null) {

  }
  if (l1 == null) {
    while (l2 != null) {
      if (carry) {
        var temp = l2.val + (carry ? 1 : 0);
        if (temp >= 10) {
          carry = 1;
          remainder = temp % 10;
        } else {
          carry = 0;
          remainder = temp;
        }
        var add = new ListNode(remainder);
        result.next = add;
        result = result.next;
        l2 = l2.next;
      } else {
        var add = new ListNode(l2.val);
        result.next = add;
        result = result.next;
        l2 = l2.next;
      }
    }
  } else {
    while (l1 != null) {
      if (carry) {
        var temp = l1.val + (carry ? 1 : 0);
        if (temp >= 10) {
          carry = 1;
          remainder = temp % 10;
        } else {
          carry = 0;
          remainder = temp;
        }
        var add = new ListNode(remainder);
        result.next = add;
        result = result.next;
        l1 = l1.next;
      } else {
        var add = new ListNode(l1.val);
        result.next = add;
        result = result.next;
        l1 = l1.next;
      }
    }
  }
  if (carry) {
    var add = new ListNode(1);
    result.next = add;
    result = result.next;
    carry = 0;
  }
  return head.next;
};