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java算法之链表两数相加

java算法之链表两数相加

作者: 上下而求索 | 来源:发表于2018-11-13 21:11 被阅读0次

    /***

    • 给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。 你可以假设除了数字 0
    • 之外,这两个数字都不会以零开头。 示例: 输入:(2 -> 4 -> 3) + (5 -> 6 -> 4) 输出:7 -> 0 -> 8 原因:342 +
    • 465 = 807
    • @author 96414

    */

    class ListNode {
        public int val;
        public ListNode next;
    }
    
    public class ListNodeTest {
        ListNode head; // 链表的表头指针
    
        public ListNode init() {
            head = new ListNode();
            head.next = null;
            return head;
        }// 初始化
    
        public void add(int x) { // 将x加入升序链表
            ListNode pre, p, q;
    //      for (pre = head, p = head.next; p != null; p = p.next, pre = pre.next)
    //          if (p.val > x)
    //              break;
            pre = head;
            p = head.next;
            q = new ListNode();
            q.val = x;
            q.next = p;
            pre.next = q; // 将q插入到pre和p之间
        }
    
        public ListNode find(int x) {// 在表中重找x,找到则返回其前驱结点的指针,找不到则返回null
            ListNode pre, p;
            pre = head;
            p = head.next;
            while (p != null && p.val != x) {
                pre = pre.next;
                p = p.next;
            }
            if (p == null)
                return null;
            return pre;
        }
    
        public void del(int x) {// 从链表中删除值为x的元素
            ListNode pre = find(x);
            if (pre == null)
                return; // 没找到
            else
                pre.next = pre.next.next; // 实施删除
        }
    
        public void showInfo() {
            for (ListNode p = head.next; p != null; p = p.next)
                System.out.print(p.val + " ");
        }
    
        public static void main(String[] args) {
            ListNodeTest test = new ListNodeTest();
            ListNodeTest test2 = new ListNodeTest(); 
            ListNode l1 = test.init();
            ListNode l2 = test2.init();
            System.out.print("请输入一组数,以-1结束:");
            Scanner sc = new Scanner(System.in);
            int x = sc.nextInt();
            while (x != -1) {
                test.add(x);
                x = sc.nextInt();
            }
            int y = sc.nextInt();
            while (y != -1) {
                test2.add(y);
                y = sc.nextInt();
            }
            System.out.print("有序链表l1为:");
            test.showInfo();
            System.out.println();
            System.out.print("有序链表l2为:");
            test2.showInfo();
            System.out.println();
            System.out.println("结果为:");
            addTwoNumbers(l1.next, l2.next);
        }
        public static void addTwoNumbers(ListNode l1, ListNode l2) {
            ListNode dummyHead = new ListNode();
            dummyHead.next=null;
            dummyHead.val=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;
                System.out.println(p.val);
                System.out.println(q.val);
                int sum = carry + x + y;
                carry = sum / 10;
                curr.next = new ListNode();
                curr.next.val = sum % 10;
                curr = curr.next;
                if (p != null) p = p.next;
                if (q != null) q = q.next;
            }
            if (carry > 0) {
                curr.next = new ListNode();
                curr.next.val = carry;
            }
            for (ListNode x1 = dummyHead.next; x1 != null; x1 = x1.next) {
                System.out.print(x1.val + " ");
            }
        }
    }
    

    参考:https://leetcode-cn.com/problems/add-two-numbers/
    https://liuyanzhao.com/2230.html

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