LeeCode（No2 - Add Two Numbers）

LeeCode是一个有意思的编程网站，主要考察程序员的算法

第二题：

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Explanation: 342 + 465 = 807.

算是一个中度难度的编程题，刚开始考虑转化为整数来加和，但任何类型的整数其实都是有上限的，所以此方法舍弃。

最后自己提交的source如下，想法也是中规中矩吧

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

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

 * }

 */

class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

      if(l1 == null || l2 == null) {

            return null;

        }

        // 为了不影响传入参数，设定游标

        ListNode cursor1 = l1;

        ListNode cursor2 = l2;

        // 结果

        ListNode result = new ListNode(0);

        ListNode cursorResult = result;

        // 进位

        int carry = 0;

        // 每位加和

        int sum;

        // 暂存游标的值

        int cursorVal1 = cursor1.val;

        int cursorVal2 = cursor2.val;

        do{

            sum = cursorVal1 + cursorVal2 + carry;

            // 是否有进位

            if(sum >= 10) {

                carry = 1;

            } else {

                carry = 0;

            }

            cursorResult.next = new ListNode(sum % 10);

            cursorResult = cursorResult.next;

            // 如果其中一个游标为空，也就是是说有链表已经遍历完

            if(cursor1 != null && cursor1.next != null) {

                cursor1 = cursor1.next;

                cursorVal1 = cursor1.val;

            } else {

                cursor1 = null;

                cursorVal1 = 0;

            }

            if(cursor2 != null && cursor2.next != null) {

                cursor2 = cursor2.next;

                cursorVal2 = cursor2.val;

            } else {

                cursor2 = null;

                cursorVal2 = 0;

            }

        } while (cursor1 != null || cursor2 != null);

        // 最后一个进位为算进去的话

        if(carry == 1) {

            cursorResult.next = new ListNode(1);

        }

        // 去掉首个元素

        return result.next;

    }

}

提交后性能表现结果如下：

虽然击败了93%的人，但是还有7%的人性能优于我的，看了下官方给出的答案，整体思路是一样的，而且它将x,y的声明放在循环内部。这可能也是统计的不精确吧

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

    ListNode dummyHead = new ListNode(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;

        int sum = carry + x + y;

        carry = sum / 10;

        curr.next = new ListNode(sum % 10);

        curr = curr.next;

        if (p != null) p = p.next;

        if (q != null) q = q.next;

    }

    if (carry > 0) {

        curr.next = new ListNode(carry);

    }

    return dummyHead.next;

}

时间复杂度和空间复杂度如下：

Complexity Analysis

Time complexity : O(\max(m, n))O(max(m,n)). Assume that mm and nn represents the length of l1l1 and l2l2 respectively, the algorithm above iterates at most \max(m, n)max(m,n) times.
Space complexity : O(\max(m, n))O(max(m,n)). The length of the new list is at most \max(m,n) + 1max(m,n)+1.

参考：

https://leetcode.com/problems/add-two-numbers

巴特西

LeeCode（No2 - Add Two Numbers）

最新文章

热门文章