2 add two numbers

---

title: Add Two Numbers
tags:
- add-two-numbers
- No.2
- medium
- list
- naive

---

Problem

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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Corner Cases

• at least one empty

input:  [], []
return: []

input:  [9], []
return: [9]

• carry with different lengths

input:  [9, 9], [9, 9, 9, 9, 9]
output: [1, 0, 0, 0, 0, 9, 8]

Solution

Naive

Pay attention to:

1. Carry. One more node may occur.
2. Initialization.
3. Numbers with different sizes.

Running time: O(n).

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

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        // O(n)

        ListNode p1 = l1;
        ListNode p2 = l2;
        ListNode h  = null;
        ListNode p  = h;

        // initialization
        int pval  = 0;
        int carry = 0;

        boolean e1  = true;
        boolean e2  = true;

        // O(n)
        // `x==null` for input:
        // `l1==null` and `l2==null`
        while (p1 != null || p2 != null) {
            // check if p1 and p2 ends
            e1 = e1 && (p1 != null);
            e2 = e2 && (p2 != null);

            // update p1, p2
            // calculate value
            if (e1 && e2) {
                pval = p1.val + p2.val + carry;
                p1   = p1.next;
                p2   = p2.next;
            }
            else if (e1 && !e2) {
                pval = p1.val + carry;
                p1   = p1.next;
            }
            else if (!e1 && e2) {
                pval = p2.val + carry;
                p2   = p2.next;
            }
            else { /* which is impossible */ }

            // update value
            carry = (pval > 9) ? 1 : 0;
            pval  = pval - carry * 10;

            // update p
            if (p == null) {
                h = new ListNode(pval);
                p = h;
            }
            else {
                p.next = new ListNode(pval);
                p      = p.next;
            }
        }

        if (carry == 1) {
            p.next = new ListNode(1);
        }

        return h;
    }
}

If in natural order

Input:  [3, 4, 2]
        [5, 6, 4]
Return: [8, 0, 7]

It's impossible to know the highest bit without traversing the whole list. Thus O(n) time of pre-processing is necessary. We can reverse the list first.

For list reversing, we set 3 pointers:a, b, c:

1. initialization a, b = null; c = head;

[a,b]  [c] -> [x] -> [x] -> [x]

1. the list has been reversed in i steps, and there is a hole between pointer b and c.

[x] <- [a] <- [b]  [c] -> [x] -> [x]

The original list is broken into 2 lists, with b and c as their heads independently.

2. a = b

[x] <- [x] <- [a,b]  [c] -> [x] -> [x]

3. b = c

[x] <- [x] <- [a]  [b,c] -> [x] -> [x]

4. c = c.next now the two heads are a and b.

[x] <- [x] <- [a]  [b] -> [c] -> [x]

5. b.next = a reverse the pointer of b and then we return to state [1].

[x] <- [x] <- [a] <- [b]  [c] -> [x]

6. finalization c == null; return b;

[x] <- [x] <- [a] <- [b]  [c]

Then the rest of the algorithm is generally same with the reversed one. Note that we change the node pointing backwards to get a natural result:

1. [y]      [p,x] -> [x] -> [x]
2. [y]   -> [p,x] -> [x] -> [x]
3. [p,y] -> [x]   -> [x] -> [x]

Running time is still O(n), with more loops:

class Solution {
    Solution () {}

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        // O(n)
        ListNode r1 = reverse(l1);
        ListNode r2 = reverse(l2);

        ListNode p1 = r1;
        ListNode p2 = r2;

        if (p1 == null && p2 == null) {return null;}

        // initialization
        int pval  = p1.val + p2.val;
        int carry = (pval > 9) ? 1 : 0;
        pval      = pval - carry * 10;

        ListNode p  = new ListNode(pval);
        p1 = p1.next;
        p2 = p2.next;

        boolean e1  = true;
        boolean e2  = true;

        // O(n)
        while (p1 != null || p2 != null) {
            // check if p1 and p2 ends
            e1 = e1 && (p1 != null);
            e2 = e2 && (p2 != null);

            // update p1, p2
            // calculate value
            if (e1 && e2) {
                pval = p1.val + p2.val + carry;
                p1   = p1.next;
                p2   = p2.next;
            }
            else if (e1 && !e2) {
                pval = p1.val + carry;
                p1   = p1.next;
            }
            else if (!e1 && e2) {
                pval = p2.val + carry;
                p2   = p2.next;
            }
            else { /* which is impossible*/ }

            // update value
            carry = (pval > 9) ? 1 : 0;
            pval  = pval - carry * 10;

            // update node
            ListNode np = new ListNode(pval);
            np.next = p;
            p       = np;
        }

        if (carry == 1) {
            ListNode np = new ListNode(1);
            np.next = p;
            p       = np;
        }

        return p;
    }

    public ListNode reverse(ListNode l) {
        ListNode a = null;
        ListNode b = null;
        ListNode c = l;

        while (c != null) {
            a      = b;
            b      = c;
            c      = c.next;
            b.next = a;
        }

        return b;
    }
}

转载于:https://www.jianshu.com/p/eae89e662450