Add Two Numbers

[zhuangjinxin]

题目： 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.

[zhuangjinxin]

解答：

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

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

时间复杂度：O(max(m,n)) 空间复杂度：O(max(m,n))

[zhuangjinxin]

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode head = new ListNode(0);
    ListNode current = head;
    int carry = 0;
    int value = 0;
    int index = 1;
    while(l1!=null || l2!=null){
        int v1 = (l1!= null)? l1.val:0;
        int v2 = (l2!=null)?l2.val:0;
        int sum = v1+v2+carry;
        carry = sum/10;
        value = sum%10;
        current.next = new ListNode(value);
        current = current.next;
        if (l1!=null) l1 =l1.next;
        if (l2!=null) l2 = l2.next;
    }
    if (carry>0){
        current.next = new ListNode(carry);
    }
    return head.next;
}