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Leetcode 002 Add two numbers

Posted on In Valine:

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.

Intuition:

这个题目总体思路非常直观。就是模拟小学的笔算来做加法。其中解题时要注意两个list长度不一样的情况（比如两位数+3位数），这种情况下直接在位数较短的那一方（先访问到null的那一方）填0就行了。

还有一个要注意的只是对list中的null如何处理。因为处理的表达式较容易，用？：表达式比if else要更方便。

Solution:

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	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
	ListNode dummyHead = new ListNode(0);
	ListNode p = l1, q = l2, curr = dummyHead;
	int carry = 0;
	//这里对于两个list长度不一致的处理方式值得学习
	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;

}