LC.P445[两数相加II]

方法一：栈+迭代

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
/**
 * 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) {
        Deque<Integer> s1 = new ArrayDeque<>(), s2 = new ArrayDeque<>();
        while (l1 != null) {
            s1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            s2.push(l2.val);
            l2 = l2.next;
        }
        ListNode head = new ListNode();
        int c = 0;
        while (!s1.isEmpty() || !s2.isEmpty()) {
            if (!s1.isEmpty()) c += s1.pop();
            if (!s2.isEmpty()) c += s2.pop();
            head.next = new ListNode(c % 10, head.next);
            c /= 10;
        }
        if (c > 0) {
            head.next = new ListNode(c, head.next);
        }
        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：$O(m+n)$

方法二：反转链表(迭代)+迭代

  1. 将$l1$反转
  2. 将$l2$反转
  3. 调用LC.P2[两数相加]，得到$l3$
  4. 反转链表$l3$，得到答案
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
/**
 * 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) {
        l1 = reverse(l1);
        l2 = reverse(l2);
        ListNode l3 = addTwo(l1, l2);
        return reverse(l3);
    }

    private ListNode addTwo(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(), tail = head;
        int c = 0;
        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;
    }

    private ListNode reverse(ListNode head) {
        ListNode cur = head, pre = null;
        while (cur != null) {
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}
  • 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：$O(1)$

方法三：反转链表(递归)+迭代

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
/**
 * 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) {
        l1 = reverse(l1, null);
        l2 = reverse(l2, null);
        ListNode l3 = addTwo(l1, l2);
        return reverse(l3, null);
    }

    private ListNode addTwo(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(), tail = head;
        int c = 0;
        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;
    }

    private ListNode reverse(ListNode cur, ListNode pre) {
        if (cur == null) return pre;
        ListNode head = reverse(cur.next, cur);
        cur.next = pre;
        return head;
    }
}
  • 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：$O(max(m,n))$