LC.P2[两数相加]

方法一：递归

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return dfs(l1, l2, 0);
    }

    private ListNode dfs(ListNode l1, ListNode l2, int c) {
        if (l1 == null && l2 == null) {
            // 如果进位了，就额外创建一个节点
            return c != 0 ? new ListNode(c) : null;
        }
        if (l1 == null) {
            l1 = l2;
            l2 = null;
        }
        c += l1.val + (l2 != null ? l2.val : 0);
        l1.val = c % 10;
        l1.next = dfs(l1.next, (l2 != null ? l2.next : null), c / 10);
        return l1;
    }
}
  • 时间复杂度：$O(max(m,n))$，其中$m$为$l1$的长度，$n$为$l2$的长度
  • 空间复杂度：$O(max(m,n))$

方法二：迭代

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(), tail = head;
        int c = 0; // 进位
        while (l1 != null || l2 != null) {
            if (l1 != null) c += l1.val;
            if (l2 != null) c += l2.val;
            tail.next = new ListNode(c % 10);
            tail = tail.next;
            c /= 10;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        if (c > 0) {
            tail.next = new ListNode(c);
        }
        return head.next;
    }
}
  • 时间复杂度：$O(max(m,n))$，其中$m$为$l1$的长度，$n$为$l2$的长度
  • 空间复杂度：$O(1)$