【栈】445. 两数相加 II（头插法）

一、题目描述

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.

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

二、题解

方法一：栈

• 由于链表给的是顺序数字，加法需要从后往前。所以需要把每一位数字拿出来先。

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    int ca = 0;
    Stack<Integer> st1 = new Stack<>();
    Stack<Integer> st2 = new Stack<>();
    ListNode t1 = l1, t2 = l2;
    while (t1 != null) {
        st1.push(t1.val);
        t1 = t1.next;
    }
    while (t2 != null) {
        st2.push(t2.val);
        t2 = t2.next;
    }
    ListNode head = null;
    while (!st1.isEmpty() || !st2.isEmpty() || ca > 0) {
        int n1 = st1.isEmpty() ? 0 : st1.pop();
        int n2 = st2.isEmpty() ? 0 : st2.pop();
        int sum = n1 + n2 + ca;
        ListNode t = new ListNode(sum%10);
        t.next = head;
        head = t;
        ca = sum / 10;
    }
    return head;
}

复杂度分析

• 时间复杂度： $O(n)$，
• 空间复杂度： $O(n)$，