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题目描述

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

代码

Approach #1

Elementary Math Intuition Keep track of the carry using a variable and simulate digits-by-digits sum starting from the head of list, which contains the least-significant digit.
Complexity Analysis
• Time complexity : O(max(m, n)). Assume that m and n represents the length of l1 and l2 respectively, the algorithm above iterates at most max(m,n) times.
• Space complexity : O(max(_m,n)). The length of the new list is at most max(m,_n)+1.
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode p = l1, q = l2;
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 (q != null) p = p.next;
if (q != null) q = q.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}