Leetcode445. Add Two Numbers II

Posted 2020-10-25 于淼

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

Tips：给定两个单链表，将两个单链表中的值相加。

思路：链表都是按照从前到后连接的，而整数加法需要从低位开始相加，即链表的后面开始相加。

我才用两个list来保存链表的val，按照从后向前的顺序相加，并将结果保存在一个list中。最后将list按照从后向前的顺序，赋值给一个新的链表。

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        while (l1 != null) {
            list1.add(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            list2.add(l2.val);
            l2 = l2.next;
        }
        int len1 = list1.size();
        int len2 = list2.size();
        //将较长的链表保存在list1中
        if (len1 < len2) {
            List<Integer> temp = list2;
            list2 = list1;
            list1 = temp;
        }
        int len = len1 > len2 ? len1 : len2;// 最后链表的长度为len或者len+1
        System.out.println(len);
        List<Integer> list = new ArrayList<>();
        int diff = Math.abs(len1 - len2);
        int tag = 0;
        //从后向前进行整数加法。
        for (int i = len - 1; i >= 0; i--) {
            //如果两个链表长度不相等，保证右对齐  ，先将能对其的位置相加.剩下的位置只加list1即可
            int temp = tag + list1.get(i);
            if (i - diff >= 0) {
                temp += list2.get(i - diff);
            }
            if (temp >= 10) {
                //进1
                tag = 1;
                temp -= 10;
            } else {
                tag = 0;
            }
            list.add(temp);
        }
        if (tag > 0)
            list.add(1);
        //打印list中的值
        for (int i = 0; i < list.size(); i++) {
            System.out.println(list.get(i));
        }
        ListNode head = new ListNode(-1);
        ListNode node = new ListNode(0);
        head.next = node;
        //将list从后向前 赋值给链表结点
        for (int i = list.size() - 1; i >= 0; i--) {
            node.next = new ListNode(list.get(i));
            node = node.next;
        }
        node.next = null;
        return head.next.next;
    }

测试代码：

public static void main(String[] args) {
        ListNode node1 = new ListNode(7);
        ListNode node2 = new ListNode(2);
        ListNode node3 = new ListNode(4);
        ListNode node4 = new ListNode(3);
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        ListNode nu = null;
        node4.next = nu;

        ListNode node11 = new ListNode(5);
        ListNode node21 = new ListNode(6);
        ListNode node31 = new ListNode(4);
        node11.next = node21;
        node21.next = node31;
        ListNode nu1 = null;
        node31.next = nu1;
        L445AddTwoNumbersII l445 = new L445AddTwoNumbersII();
        ListNode head = l445.addTwoNumbers(node11, node1);
        while (head != null) {
            System.out.println("~~~~~~~~~~~~");
            System.out.println(head.val);
            head = head.next;
        }
}