卡在Leetcode的案例3“两个和相加”

Question

目前我每天都在做Leetcode streak，我不知道为什么我的代码无法通过案例3。（我已经完成了修改和跟踪代码，但也许我有点笨，无法通过）没有发现错误。）希望这不是任何愚蠢的傻瓜错误，感谢您的帮助！

参考这个leetcode问题。

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

 *     ListNode(int val, ListNode next) { 
            this.val = val; 
            this.next = next; 
    }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        // Making an empty LinkList to store answers
        ListNode answer = new ListNode();

        // A pointer the transverse the answer
        ListNode current = answer;

        // Dont forget carries
        int carry = 0;

        // Setting up global variable
        int l1_value, l2_value;

        // If carry = 0 means Answer is DONE!
        while(l1 != null || l2 != null || carry != 0) {
            // If l1 -> null, then 0
            if (l1 != null) {
                l1_value = l1.val;              // Directly get val if it is public
            } else l1_value = 0;

            if (l2 != null) {
                l2_value = l2.val;
            } else l2_value = 0;

            // DONT FORGET ABOUT CARRY
            int total = l1_value + l2_value + carry;
            int remainder = total % 10;


            current.next = new ListNode(remainder, current.next);
            carry = total / 10;

            // Transverse to next, else null
            if (l1 != null && l1.next != null) {
                l1 = l1.next;
            } else l1 = null;

            
            // Error occurs because at some point this would become NULL, causes the NullPointerException
            if (l2 != null && l2.next != null) {
                l2 = l2.next;
            } else l2 = null;
            
            // I think the extra 0s, 1s are from the CARRY here [Added if condition]
            // Error still occur but much better, added [carry != 0], T1/2 done! but T3 zzzz
            if (l1 == null && l2 == null && carry != 0) {
                current.next = new ListNode(carry, current.next);
            }
            current = current.next;
        }
        return answer.next;
    }
}

问题与解答：

l1 = [9,9,9,9,9,9,9]
l2 = [9,9,9,9]

预期输出 = [8,9,9,9,0,0,0,1]
我的输出= [8,9,9,9,0,0,1,1,0]

Answer 1

你基本上已经自己解决了。问题正是您所期望的：

        // I think the extra 0s, 1s are from the CARRY here [Added if condition]
        // Error still occur but much better, added [carry != 0], T1/2 done! but T3 zzzz
        if (l1 == null && l2 == null && carry != 0) {
            current.next = new ListNode(carry, current.next);
        }

这只是多余的。您之前已经添加了进位：

int total = l1_value + l2_value + carry;

所以如果你有进位，总计总是 != 0。因此余数是 != 0 因此，这个 ListNode 就足够了：

current.next = new ListNode(remainder, current.next);

只需删除ListNode的额外创建即可，结果将如预期。

我的完整代码：

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        // Making an empty LinkList to store answers
        ListNode answer = new ListNode();

        // A pointer the transverse the answer
        ListNode current = answer;

        // Dont forget carries
        int carry = 0;

        // Setting up global variable
        int l1_value, l2_value;

        // If carry = 0 means Answer is DONE!
        while (l1 != null || l2 != null || carry != 0) {
            // If l1 -> null, then 0
            if (l1 != null) {
                l1_value = l1.val;              // Directly get val if it is public
            } else l1_value = 0;

            if (l2 != null) {
                l2_value = l2.val;
            } else l2_value = 0;

            // DONT FORGET ABOUT CARRY
            int total = l1_value + l2_value + carry;
            int remainder = total % 10;


            current.next = new ListNode(remainder, current.next);
            carry = total / 10;

            // Transverse to next, else null
            if (l1 != null && l1.next != null) {
                l1 = l1.next;
            } else l1 = null;


            // Error occurs because at some point this would become NULL, causes the NullPointerException
            if (l2 != null && l2.next != null) {
                l2 = l2.next;
            } else l2 = null;
            
            current = current.next;
        }
        return answer.next;
    }

    public static void main(String[] args) {
        var l1 = new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9)))))));
        var l2 = new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9))));


        ListNode result = new Solution().addTwoNumbers(l1, l2);

        print(result);
    }

    private static void print(ListNode result) {
        System.out.print(result.val);

        if (result.next != null) {
            print(result.next);
        }
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode() {
    }

    ListNode(int val) {
        this.val = val;
    }

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}