LeetCodeOJ_002: Add Two Numbers

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原题目：https://leetcode.com/problems/add-two-numbers/

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题目

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Total Accepted: 293186
Total Submissions: 1071927
Difficulty: Medium
Contributor: LeetCode

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

AC代码

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int carry = 0;        ListNode dummy = new ListNode(-1);        ListNode cur = dummy;        while (l1 != null || l2 != null) {            int sum = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;            carry = sum / 10;            cur.next = new ListNode(sum % 10);            cur = cur.next;            l1 = l1 == null ? null : l1.next;            l2 = l2 == null ? null : l2.next;        }        if (carry != 0) cur.next = new ListNode(carry);        return dummy.next;    }}

Complexity Analysis

• Time complexity : O(max(m, n)). Assume that mm 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 .

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这是我的AC代码

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {       if(l1 == null && l2 == null) {            return null;        }        ListNode head = new ListNode(0);        ListNode point = head;        int carry = 0;        while(l1 != null && l2!=null){            int sum = carry + l1.val + l2.val;            point.next = new ListNode(sum % 10);            carry = sum / 10;            l1 = l1.next;            l2 = l2.next;            point = point.next;        }        while(l1 != null) {            int sum =  carry + l1.val;            point.next = new ListNode(sum % 10);            carry = sum /10;            l1 = l1.next;            point = point.next;        }        while(l2 != null) {            int sum =  carry + l2.val;            point.next = new ListNode(sum % 10);            carry = sum /10;            l2 = l2.next;            point = point.next;        }        if (carry != 0) {            point.next = new ListNode(carry);        }        return head.next;     }}