LeetCode2_Add Two Numbers

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My answer

1. 第二遍提交

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if (l1 == null) {            return l2;        }        if (l2 == null) {            return l1;        }        ListNode front = null, rear = null;        int sum = 0;        while (true) {            if (l1 == null && l2 == null) {                if (sum == 1) {                    rear.next = new ListNode(1);                    rear = rear.next;                }                break;            }            // add two numbers            if (l1 != null) {                sum += l1.val;                l1 = l1.next;            }            if (l2 != null) {                sum += l2.val;                l2 = l2.next;            }            // sum is greater than 10            ListNode n = null;            if (sum >= 10) {                n = new ListNode(sum - 10);                sum = 1;            } else {                n = new ListNode(sum);                sum = 0;            }            // LinkedList addRear            if (front == null) {                front = n;                rear = front;            } else {                rear.next = n;                rear = rear.next;            }        }        return front;    }

2. 第一遍提交

    /**     * 该方法可以写的更优雅：1) 多余的变量 ptr1,ptr2; 2) sum和flag可以合并     */    public ListNode addTwoNumbers_version0(ListNode l1, ListNode l2) {        if (l1 == null) {            return l2;        }        if (l2 == null) {            return l1;        }        ListNode ptr1 = l1, ptr2 = l2;        ListNode front = null, rear = null;        boolean flag = false; // add one to the next bit        int sum = 0;        while (true) {            if (ptr1 == null && ptr2 == null) {                if (flag) {                    rear.next = new ListNode(1);                    rear = rear.next;                }                break;            }            if (flag) {                sum = 1;            } else {                sum = 0;            }            // add two numbers            if (ptr1 != null) {                sum += ptr1.val;                ptr1 = ptr1.next;            }            if (ptr2 != null) {                sum += ptr2.val;                ptr2 = ptr2.next;            }            // sum is greater than 10            if (sum >= 10) {                sum = sum - 10;                flag = true;            } else {                flag = false;            }            // LinkedList addRear            ListNode n = new ListNode(sum);            if (front == null) {                front = n;                rear = front;            } else {                rear.next = n;                rear = rear.next;            }        }        return front;    }

Summary

1. Keys:

1.1 Two linked lists' lengths are not equal

1.2 Two digit sum (especially the final bit) is more than 10.

1.3 AddRear method of the Linked List

2. 加法运算涉及到进位的问题, thus the digits are stored in reverse order.

3. Submission Details

第一遍提交

尽管绝对的运行时间跟提交那时机器负载有关，但这个图很有意思！！

1. 大量样本下，程序员的能力呈正态分布。

2. Java语言最慢，C++最快！！

3. 第一遍，我的编程水平（if 大家都是自己做并且一次通过）属于中偏下，代码执行时间还是太慢，值得再优化，打磨。

4. 第二遍提交runtime 提升到了435ms.

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