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addtwonumbers

admin 11月 13, 2020 0

1.Description

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

2.Solution

一开始边界条件没想清楚，没确定什么时候能够结束循环（什么时候进位生成新的节点）

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode listNode = new ListNode(0);
    ListNode result = listNode;
    int carry = 0;
    while (l1 != null || l2 != null) {
        boolean flag = false; //标识是否应该退出循环
        int val = l1.val + l2.val + carry;
        carry = val >= 10 ? 1 : 0;
        listNode.val = val % 10;
        if (l1.next != null || l2.next != null || carry == 1) {
            flag = true;
            listNode.next = new ListNode(0);
            listNode = listNode.next;
        }
        if (l1.next != null) {
            flag = true;
            l1 = l1.next;
        } else {
            l1.val = 0;
        }
        if (l2.next != null) {
            flag = true;
            l2 = l2.next;
        } else {
            l2.val = 0;
        }
        if (!flag) {
            break;
        }
    }
    return result;
}

3.其他

时间复杂度：O（max(m,n)） m&n代表l1&l2的长度
空间复杂度：O（max(m,n)）新列表最长为max(m,n)+1
参考答案：

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;
}