链表逆序 从尾到头打印节点
2018-04-16更新: 递归逆序链表
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
|
class LinkNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
def __str__(self):
if self.next is not None:
return '{},{}'.format(self.val, self.next)
return '{}'.format(self.val)
def reverse(cur, pre=None):
next = cur.next
cur.next = pre
if next is None:
return cur
else:
return reverse(next, cur)
if __name__ == '__main__':
node = LinkNode(1, LinkNode(2, LinkNode(3, LinkNode(4))))
print(node)
head = reverse(node)
print(head)
|
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
|
class ListNode {
int val;
ListNode next;
}
/**
* 链表逆序
* 使用三个指针: pre, cur, next 完成
*/
public ListNode reverse(ListNode node) {
if (node == null) {
return null;
}
ListNode cur = node;
ListNode pre = null;
while (cur != null) { // 当前节点为 null 说明遍历结束了
ListNode next = cur.next; // 暂存当前节点的下一个节点
cur.next = pre; // 当前节点的 next 指向前一个节点
pre = cur; // 前一个节点移到当前节点
cur = next; // 当前节点移到下一个节点
}
return pre;
}
|
一次操作过程:
next=cur.next;
pre cur next
null [] -> [] -> [] -> [] -> null
cur.next = pre;
pre cur next
null <-[] x [] -> [] -> [] -> null
pre = cur; cur = next;
pre cur
next
null <-[] x [] -> [] -> [] -> null
1
2
3
4
5
6
7
8
9
10
11
12
|
//链表逆序(使用栈做辅助)
public ListNode reverseList(ListNode head) {
Stack<ListNode> stack = new Stack<>();
while (head != null) {
stack.push(head);
head = head.next;
}
while (!stack.isEmpty()) {
head = stack.pop();
}
return head;
}
|
1
2
3
4
5
6
7
8
9
10
11
12
|
//从尾到头打印节点(使用栈做辅助)
public ArrayList<Integer> reList(ListNode head) {
ArrayList<Integer> arrayList = new ArrayList<>();
Stack<Integer> stack = new Stack<>();
while (head != null) {
stack.push(head.val);
}
while (!stack.isEmpty()) {
arrayList.add(stack.pop());
}
return arrayList;
}
|
两个栈实现队列的功能
public class Solution {
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>();
public void push(int node) {
stack1.push(node);
}
public int pop() {
if(!stack2.isEmpty()){
return stack2.pop();
}
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
return stack2.pop();
}
}
