本文共 3525 字，大约阅读时间需要 11 分钟。

//给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位// 数字。//// 如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。 //// 您可以假设除了数字 0 之外，这两个数都不会以 0 开头。 //// 示例： //// 输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)//输出：7 -> 0 -> 8//原因：342 + 465 = 807// // Related Topics 链表 数学

package com.hunter.leetcode;/** * @author hunter.yang * @version 1.0 * @description null * @date 2020/10/25 16:28 */public class Solution2 {        public static class ListNode {        int val;        ListNode next;        ListNode() {}        ListNode(int val) { this.val = val; }        ListNode(int val, ListNode next) { this.val = val; this.next = next; }    }    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode result = new ListNode();        ListNode preNode = new ListNode();        boolean ifNeedAddOne = false;        boolean firstCal = true;        while (null != l1 || null != l2) {            ListNode node = new ListNode();            int l1Value = 0;            int l2Value = 0;            if (null != l1) {                l1Value = l1.val;            }            if (null != l2) {                l2Value = l2.val;            }            if (ifNeedAddOne) {                if (null == l1) {                    l2Value = l2Value + 1;                    node.val = gtTenLeftValue(0, l2Value);                } else if (null == l2){                    l1Value = l1Value + 1;                    node.val = gtTenLeftValue(l1Value, 0);                } else {                    l1Value = l1Value + 1;                    node.val = gtTenLeftValue(l1Value, l2Value);                }            } else {                if (null == l1) {                    node.val = gtTenLeftValue(0, l2Value);                } else if (null == l2){                    node.val = gtTenLeftValue(l1Value, 0);                } else {                    node.val = gtTenLeftValue(l1Value, l2Value);                }            }            ifNeedAddOne = judge(l1Value, l2Value);            if(firstCal) {                result = node;                firstCal = false;                preNode = result;            } else {                preNode.next = node;                preNode = node;            }            if (null != l1) {                l1 = l1.next;            }            if (null != l2) {                l2 = l2.next;            }            if (null == l1 && null == l2 && ifNeedAddOne) {                ListNode lastNode = new ListNode();                lastNode.val = 1;                preNode.next = lastNode;            }        }        return result;    }    // 判断两个数相加是否需要向前一位加1    private static boolean judge(int nodeValue1, int nodeValue2) {        if (nodeValue1 + nodeValue2 >= 10) {            return true;        } else {            return false;        }    }    private static int gtTenLeftValue(int nodeValue1, int nodeValue2) {        int sum = nodeValue1 + nodeValue2;        if (sum >= 10) {            return sum%10;        } else {            return sum;        }    }        public static void main(String[] args) {        ListNode l1 = new ListNode(9,                new ListNode(9,                        new ListNode(9,                                new ListNode(9,                                        new ListNode(9,                                                new ListNode(9,                                                        new ListNode(9)))))));        ListNode l2 = new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9))));        System.out.println(addTwoNumbers(l1, l2));    }}

转载地址：http://bzugf.baihongyu.com/