LeetCode--Add Two Numbers

最新推荐文章于 2024-08-12 10:14:14 发布

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本文介绍了一种解决链表加法问题的方法，通过遍历两个非空链表，将其逆序排列成数字相加，再将结果逆序排列成链表输出。文章详细描述了算法实现过程，包括节点值的两两相加、进位处理及最终链表的构建。

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

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.

这道题简单讲就是输入两个链表，将其逆序排列成数字后相加，并将结果逆序排列成链表输出

我的想法是：

先将两个链表从第一个开始两两相加，放入一个链表中
将链表转化为给定格式返回

代码见：https://github.com/Wen-Xueliang/LeetCode/blob/master/src/main/java/eng/lab/leetcode/algorithms/TwoSum.java

时间与空间使用信息

Runtime: 4 ms, faster than 8.40% of Java online submissions for Add Two Numbers.

Memory Usage: 45 MB, less than 84.01% of Java online submissions for Add Two Numbers.

public class AddTwoNumbers {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = null;
        ListNode currentNode1 = l1;
        ListNode currentNode2 = l2;
        List<Integer> list = new ArrayList<Integer>();
        int carry = 0;

        while(true) {
            int node1Value = 0;
            int node2Value = 0;
            if(currentNode1 == null && currentNode2 == null) {
                if(carry == 1) list.add(carry);
                break;
            } else if(currentNode1 == null){
                node2Value = currentNode2.val;
                currentNode2 = currentNode2.next;
            } else if(currentNode2 == null){
                node1Value = currentNode1.val;
                currentNode1 = currentNode1.next;
            } else {
                node1Value = currentNode1.val;
                node2Value = currentNode2.val;
                currentNode1 = currentNode1.next;
                currentNode2 = currentNode2.next;
            }
            int sum = node1Value + node2Value + carry;

            if(sum > 9) {
                carry = 1;
                sum -= 10;
            } else {
                carry = 0;
            }
            list.add(sum);
        }
        for(int i = list.size() - 1; i >= 0; i --) {
            ListNode currentNode = null;
            if(result == null) {
                result = new ListNode(list.get(i));
            } else {
                currentNode = new ListNode(list.get(i));
                currentNode.next = result;
                result = currentNode;
            }
        }
        return result;
    }

    public static class ListNode {
        int val;
        ListNode next;
        ListNode(int x) { val = x; }
    }
}