445. 两数相加 II

题目描述

原题

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶：

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例：

输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7

题解

方法一

/**
 * 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) {
        //先翻转链表 然后两个链表相加
      //再翻转结果链表
        l1 = reverse(l1);
        l2 = reverse(l2);
        ListNode dummyNode = new ListNode();
        ListNode curr = dummyNode;
        int carry = 0;
        while(l1 != null || l2 !=null){
            int x = l1!=null ? l1.val:0;
            int y = l2!=null ? l2.val:0;
            int sum = x + y + carry;
            carry = sum / 10;
            curr.next = new ListNode(sum%10);
            curr = curr.next;
            if(l1!=null) l1 = l1.next;
            if(l2!=null) l2 = l2.next;
        }
        if(carry>0){
            curr.next = new ListNode(carry);
        }
        return reverse(dummyNode.next);
    }

    private ListNode reverse(ListNode node){
        ListNode pre = null;
        ListNode curr = node;
        while(curr!=null){
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }
}

方法二

/**
 * 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) {
        LinkedList<ListNode> stack1 = addToStack(l1);
        LinkedList<ListNode> stack2 = addToStack(l2);
        LinkedList<ListNode> stack3 = new LinkedList<>();
        int carry = 0;
        while(!stack1.isEmpty() || !stack2.isEmpty()||carry > 0){
            int x = !stack1.isEmpty() ? stack1.pop().val:0;
            int y = !stack2.isEmpty() ? stack2.pop().val:0;
            int sum = x + y + carry;
            carry = sum / 10;
            stack3.push(new ListNode(sum % 10));
        }
        ListNode dummyNode = new ListNode();
        ListNode curr = dummyNode;
        while(!stack3.isEmpty()){
            curr.next = stack3.pop();
            curr = curr.next;
        }
        return dummyNode.next;
    }
    private LinkedList<ListNode> addToStack(ListNode node){
        LinkedList<ListNode> stack = new LinkedList<>();
        while(node!=null){
            stack.push(node);
            node = node.next;
        }
        return stack;
    }
}

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        //栈里面可以直接存储值 不用存储ListNode
        Deque<Integer> stack1 = new LinkedList<Integer>();
        Deque<Integer> stack2 = new LinkedList<Integer>();
        while (l1 != null) {
            stack1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            stack2.push(l2.val);
            l2 = l2.next;
        }
        int carry = 0;
        ListNode pre = null;
        while (!stack1.isEmpty() || !stack2.isEmpty() || carry != 0) {
            int a = stack1.isEmpty() ? 0 : stack1.pop();
            int b = stack2.isEmpty() ? 0 : stack2.pop();
            int sum = a + b + carry;
            carry = sum / 10;
            sum %= 10;
            //这里借助了链表翻转的思路
            ListNode curr = new ListNode(sum);
            curr.next = pre;
            pre = curr;
        }
        return pre;
    }
}