正好复习一下链表

Question

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse orderand 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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Code provided:

/**
 * 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) {
        
    }
}

Solution

_{source: leetcode}

En

Before starting summing up two lists, create a new linked list to store the summed up value. Given $l1$ and $l2$ , starting with the head of $l1$ and $l2$ . Each digit is in the range of $0...9$ , so it is possible that summing two digits may result in a value that is greater than 9, therefore require $2$ digits to represent. Hence, extra digit is needed during each iteration to carry the overflowing value.

CN

代码中已经提供了 $l1$ 和 $l2$ ，因此所需要的步骤就是创建一个新的空的linked list。从头开始迭代 $l1$ 和 $l2$ ，并将相加的数值存入到新的linked list之中。值得注意的事情就是，每个数字的区间是 $0...9$ ，因此，当两位数字相加的时候就可能会出现十位数的可能，也就因此需要额外的变量保存十位数的数字。

Code

Java

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    
    While (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

Python3

def addTwoNumbers(self, l1, l2):
    dummy = cur = ListNode(0)
    carry = 0
    while l1 or l2 or carry:
        if l1:
            carry += l1.val
            l1 = l1.next
		if l2:
            carry += l2.val
            l2 = l2.next
        cur.next = ListNode(carry%10)
        cur = cur.next
        # floor division
        # 10 / 4 => 2.5
        # 10 // 4 = 2
        carry //= 10
    return dummy.next

Complexity Analysis

Time Complexity:
Space Complexity:

Reference

Python solution - 80ms - Easy to read

Solution