題目：Add Two Numbers

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.

Input:

• 鏈表頭結點 :: ListNode
• 另一個鏈表頭結點 :: ListNode

Output:

• 相加后鏈表頭結點 :: ListNode

Intuition:

這題就很直接的每個元素相加就好，但是tricky的地方在于如何處理進位。很顯然我們需要一個變量來存儲當前位的進位狀況然后在下一位的時候添加上。

Code:

TC: (n) SC: (1)

public ListNode addTwoNum(ListNode l1, ListNode l2){
  ListNode dummy = new ListNode(-1);
  ListNode p = dummy;
  ListNode p1 = l1;
  ListNode p2 = l2;
  int sum = 0;
  int carry = 0;

  while (p1 != null && p2 != null){
    if (p1 != null){
      sum += p1.val;
      p1 = p1.next;
    }
    
    if (p2 != null){
      sum += p2.val;
      p2 = p2.next;
    }
    //save carry for next round addition
    carry = sum / 10;
    p.next = new ListNode( sum % 10);
    p = p.next;
    sum /= 10;
  }
  if(carry == 1) p.next = new ListNode(1);
  return dummy.next;
}

題目：Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains 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 list is not allowed.

Input:

• 鏈表頭結點 :: ListNode
• 另一個鏈表頭結點 :: ListNode

Output:

• 相加后鏈表頭結點 :: ListNode

Intuition:

跟第一題不同的地方就在于這回鏈表表示的數字跟我們默認的數字是一樣的，最小位在左邊。需要解決的問題其實還是進位，不過我們得倒著加，即把左邊的高位數的數先存著，低位數加完后再來加高位數。是不是聽著挺耳熟？這不就是先進后出嘛？所以我們需要一個stack。

Code:

TC: (n) SC: (n)

public ListNode addTwoNum(ListNode l1, ListNode l2){
  Deque<Integer> stk1 = new ArrayDeque<>();
  ListNode dummy = new ListNode(-1);
  ListNode p = dummy;
  ListNode p1 = l1;
  ListNode p2 = l2;
  int sum = 0;
  int carry = 0;
  //use stack to reverse order
  while (p1 != null && p2 != null){
    if (p1 != null){
      stk1.push(p1.val);
      p1 = p1.next;
    }
    
    if (p2 != null){
      stk2.push(p2.val);
      p2 = p2.next;
    }
  }
    
  while(!stk1.isEmpty() || !stk2.isEmpty()){
    //save carry for next round addition
    if (!stk1.isEmpty()){
      sum += stk1.pop()
    }
    
    if (!stk2.isEmpty()){
      sum += stk1.pop()
    }
    carry = sum / 10;
    ListNode cur = new ListNode(carry);
    cur.next = ListNode( sum % 10);
    cur =
    p = p.next;
    sum /= 10;
  }
  return dummy.next;
}

Reference

https://leetcode.com/problems/add-two-numbers/description/
https://leetcode.com/problems/add-two-numbers-ii/discuss/