leetcode-n2-两数相加

leetcode-两数相加

下面的类,能够重用,修改一下能够运行在在线IDE,给出了测试main()

题目源地址

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/*
 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，而且它们的每一个节点只能存储 一位 数字。

 若是，咱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。

 您能够假设除了数字 0 以外，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
 缘由：342 + 465 = 807

 来源：力扣（LeetCode）
 连接：https://leetcode-cn.com/problems/add-two-numbers
 著做权归领扣网络全部。商业转载请联系官方受权，非商业转载请注明出处。
 */

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

class ListNode {
    int val;

    ListNode next;

    ListNode(int x) {
        val = x;
    }
}

class Solution {

    public static void main(String[] args) {
        Solution solution = new Solution();

        //基本测试
        ListNode res = solution.addTwoNumbers(solution.long2ListNode(342), solution.long2ListNode(465));
        sout(res);

        //特殊:单链表
        res = solution.addTwoNumbers(solution.long2ListNode(619), solution.long2ListNode(0));
        sout(res);
        //特殊:边界进位
        res = solution.addTwoNumbers(solution.long2ListNode(5), solution.long2ListNode(5));
        sout(res);

        res = solution.addTwoNumbers(solution.long2ListNode(1), solution.long2ListNode(99));
        sout(res);
    }

    private static void sout(ListNode l3) {
        while (l3 != null) {
            System.out.println(l3.val);
            l3 = l3.next;
        }
        System.out.println("---------");
    }

    /**
     * 说明: 使用进位
     * @author suwenguang
     * @date 2019/6/10
     * @return n2.ListNode <- 返回类型
     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        //参数校验
        if (l1 == null || l2 == null) {
            return null;
        }
        if (l1.val < 0 || l2.val < 0) {
            return null;
        }
        //主逻辑
        //进位链表
        ListNode carrier = new ListNode(0);
        //结果链表
        ListNode res = new ListNode(0);
        //临时链表
        ListNode temp = res;

        while (l1 != null && l2 != null) {
            //相加进位
            int val = l1.val + l2.val + carrier.val;
            if (val >= 10) {
                carrier.next = new ListNode(1);
            } else {
                carrier.next = new ListNode(0);
            }
            //赋值res
            int remain = val % 10;
            temp.next = new ListNode(remain);

            //指针移动
            temp = temp.next;
            l1 = l1.next;
            l2 = l2.next;
            carrier = carrier.next;
        }
        while (l1 != null) {
            int val = carrier.val + l1.val;
            if (val >= 10) {
                carrier.next = new ListNode(1);
            } else {
                carrier.next = new ListNode(0);
            }

            //赋值
            int remain = val % 10;
            temp.next = new ListNode(remain);

            //指针移动
            l1 = l1.next;
            carrier = carrier.next;
            temp = temp.next;

        }
        while (l2 != null) {
            int val = carrier.val + l2.val;
            if (val >= 10) {
                carrier.next = new ListNode(1);
            } else {
                carrier.next = new ListNode(0);
            }

            //赋值
            int remain = val % 10;
            temp.next = new ListNode(remain);

            //指针移动
            l2 = l2.next;
            carrier = carrier.next;
            temp = temp.next;

        }

        //断定是否还有进位
        if (carrier.val == 1) {
            temp.next = new ListNode(1);
        }

        return res.next;
    }

    /**
     * 说明:暴力解法
     * @author suwenguang
     * @date 2019/6/10
     * @return n2.ListNode <- 返回类型
     */
    @Deprecated
    public ListNode addTwoNumbersFail(ListNode l1, ListNode l2) {
        //参数校验
        if (l1 == null || l2 == null) {
            return null;
        }
        if (l1.val < 0 || l2.val < 0) {
            return null;
        }
        //将链表转为整数
        long a = listNode2Long(l1);
        long b = listNode2Long(l2);

        //相加
        long res = a + b;

        //结果转链表返回
        ListNode head = long2ListNode(res);

        return head.next;
    }

    /**
     * 说明:  ListNode 转 long
     * @author suwenguang
     * @date 2019/6/10
     * @return long <- 返回类型
     */
    private long listNode2Long(ListNode l2) {
        int i = 1;
        long res = 0;
        ListNode tempNode;
        tempNode = l2;
        while (tempNode != null) {
            res = res + (tempNode.val * i);
            tempNode = tempNode.next;
            i *= 10;
        }
        return res;
    }

    /**
     * 说明: long 转 ListNode
     * @author suwenguang
     * @date 2019/6/10
     * @return n2.ListNode <- 返回类型
     */
    private ListNode long2ListNode(long res) {
        ListNode tempNode;
        long temp = 11;

        ListNode head = new ListNode(-1);
        tempNode = head;

        while (res / 10 > 0) {
            temp = res % 10;
            ListNode next = new ListNode((int) temp);
            tempNode.next = next;
            tempNode = next;
            res /= 10;
        }
        ListNode end = new ListNode((int) res);
        tempNode.next = end;
        return head.next;
    }
}

2019-06-10 19:37:14 星期一