Question

02.05. Sum Lists

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Description

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

 

Example:

Input: (7 -> 1 -> 6) + (5 -> 9 -> 2). That is, 617 + 295.

Output: 2 -> 1 -> 9. That is, 912.

Follow Up: Suppose the digits are stored in forward order. Repeat the above problem.

Example:

Input: (6 -> 1 -> 7) + (2 -> 9 -> 5). That is, 617 + 295.

Output: 9 -> 1 -> 2. That is, 912.

Solutions

Solution 1: Simulation

We traverse two linked lists $l_1$ and $l_2$ simultaneously, and use a variable $carry$ to indicate whether there is a carry-over currently.

During each traversal, we take out the current digit of the corresponding linked list, calculate the sum of them and the carry-over $carry$, then update the value of the carry-over, and finally add the value of the current digit to the answer linked list. The traversal ends when both linked lists have been traversed and the carry-over is $0$.

Finally, we return the head node of the answer linked list.

The time complexity is $O(\max(m, n))$, where $m$ and $n$ are the lengths of the two linked lists respectively. We need to traverse all positions of the two linked lists, and it only takes $O(1)$ time to process each position. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    dummy = ListNode()
    carry, curr = 0, dummy
    while l1 or l2 or carry:
      s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
      carry, val = divmod(s, 10)
      curr.next = ListNode(val)
      curr = curr.next
      l1 = l1.next if l1 else None
      l2 = l2.next if l2 else None
    return dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    int carry = 0;
    ListNode cur = dummy;
    while (l1 != null || l2 != null || carry != 0) {
      int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
      carry = s / 10;
      cur.next = new ListNode(s % 10);
      cur = cur.next;
      l1 = l1 == null ? null : l1.next;
      l2 = l2 == null ? null : l2.next;
    }
    return dummy.next;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* dummy = new ListNode(0);
    ListNode* cur = dummy;
    int carry = 0;
    while (l1 || l2 || carry) {
      carry += (!l1 ? 0 : l1->val) + (!l2 ? 0 : l2->val);
      cur->next = new ListNode(carry % 10);
      cur = cur->next;
      carry /= 10;
      l1 = l1 ? l1->next : l1;
      l2 = l2 ? l2->next : l2;
    }
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
  dummy := &ListNode{}
  cur := dummy
  carry := 0
  for l1 != nil || l2 != nil || carry > 0 {
    if l1 != nil {
      carry += l1.Val
      l1 = l1.Next
    }
    if l2 != nil {
      carry += l2.Val
      l2 = l2.Next
    }
    cur.Next = &ListNode{Val: carry % 10}
    cur = cur.Next
    carry /= 10
  }
  return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
  if (l1 == null || l2 == null) {
    return l1 && l2;
  }
  const dummy = new ListNode(0);
  let cur = dummy;
  while (l1 != null || l2 != null) {
    let val = 0;
    if (l1 != null) {
      val += l1.val;
      l1 = l1.next;
    }
    if (l2 != null) {
      val += l2.val;
      l2 = l2.next;
    }
    if (cur.val >= 10) {
      cur.val %= 10;
      val++;
    }
    cur.next = new ListNode(val);
    cur = cur.next;
  }
  if (cur.val >= 10) {
    cur.val %= 10;
    cur.next = new ListNode(1);
  }
  return dummy.next;
}

impl Solution {
  pub fn add_two_numbers(
    mut l1: Option<Box<ListNode>>,
    mut l2: Option<Box<ListNode>>
  ) -> Option<Box<ListNode>> {
    let mut dummy = Some(Box::new(ListNode::new(0)));
    let mut cur = dummy.as_mut();
    while l1.is_some() || l2.is_some() {
      let mut val = 0;
      if let Some(node) = l1 {
        val += node.val;
        l1 = node.next;
      }
      if let Some(node) = l2 {
        val += node.val;
        l2 = node.next;
      }
      if let Some(node) = cur {
        if node.val >= 10 {
          val += 1;
          node.val %= 10;
        }
        node.next = Some(Box::new(ListNode::new(val)));
        cur = node.next.as_mut();
      }
    }
    if let Some(node) = cur {
      if node.val >= 10 {
        node.val %= 10;
        node.next = Some(Box::new(ListNode::new(1)));
      }
    }
    dummy.unwrap().next
  }
}

/**
 * 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) {
  let carry = 0;
  const dummy = new ListNode(0);
  let cur = dummy;
  while (l1 || l2 || carry) {
    carry += (l1?.val || 0) + (l2?.val || 0);
    cur.next = new ListNode(carry % 10);
    carry = Math.floor(carry / 10);
    cur = cur.next;
    l1 = l1?.next;
    l2 = l2?.next;
  }
  return dummy.next;
};