Add Two Numbers II

Question

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in forward order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

Example:

Given 6->1->7 + 2->9->5. That is, 617 + 295.

Return 9->1->2. That is, 912.

Analysis

相比于数字为倒序的链表相加问题(add two numbers),顺序的数字(单向)链表存在处理进位的问题,因为没有反向指针,所以后节点所得的进位,难以加到前一个节点上。 此外,与 add two numbers 问题一样,不能简单地将链表转为整形int或长整型long,因为很有可能超出范围,使用链表的好处正是在于可以几乎无限制地增加number的位数,如果有转换的过程,则很有可能丢失精度。

因此,最简单的解决办法,就是将 add two numbers ii 问题,转化为 add two numbers 问题:先分别对每个number链表进行反转,然后进行相加,最后将所得链表再进行反转。

复杂问题可以通过分析,转化和分解为更小的,更基础的问题。

Solution

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

    public ListNode reverse(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }

    public ListNode addLists(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) {
            return null;
        }

        ListNode head = new ListNode(0);
        ListNode pointer = head;

        int carry = 0;

        while (l1 != null && l2 != null) {

            int sum = l1.val + l2.val + carry;
            carry = sum / 10;

            pointer.next = new ListNode(sum % 10);
            pointer = pointer.next;
            l1 = l1.next;
            l2 = l2.next;
        }

        while (l1 != null) {
            int sum = l1.val + carry;
            carry = sum / 10;
            pointer.next = new ListNode(sum % 10);
            pointer = pointer.next;
            l1 = l1.next;
        }

        while (l2 != null) {
            int sum = l2.val + carry;
            carry = sum / 10;
            pointer.next = new ListNode(sum = sum % 10);
            pointer = pointer.next;
            l2 = l2.next;
        }

        if (carry != 0) {
            pointer.next = new ListNode(carry);
        }

        return head.next;
    }
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    public ListNode addLists2(ListNode l1, ListNode l2) {
        l1 = reverse(l1);
        l2 = reverse(l2);

        return reverse(addLists(l1, l2));
    }
}

results matching ""

    No results matching ""