剑指offer 06. 从尾到头打印链表
输入一个链表的头节点,从尾到头反过来返回每个节点的值(用数组返回)。
示例 1:
输入:head = [1,3,2]
输出:[2,3,1]
限制:
0 <= 链表长度 <= 10000
解法一:
利用栈先进后出的性质求解。时间复杂度O(n),空间复杂度O(n)。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
Stack<Integer> st = new Stack<>();
ListNode h = head;
int count = 0;
while(h != null){
st.push(h.val);
count++;
h = h.next;
}
int[] nums = new int[count];
for(int j=0;j<count;j++){
nums[j] = st.pop();
}
return nums;
}
}
解法二:
递归求解
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
int count ,i = 0;
int[] nums;
public int[] reversePrint(ListNode head) {
if(head != null)
count++;
else
return new int[count];
nums = reversePrint(head.next);
nums[i] = head.val;
i++;
return nums;
}
}