面试题 06. 从尾到头打印链表

题目描述

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

方法一：顺序遍历 + 反转

我们可以顺序遍历链表，将每个节点的值存入数组中，然后将数组反转。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为链表的长度。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def reversePrint(self, head: ListNode) -> List[int]:
        ans = []
        while head:
            ans.append(head.val)
            head = head.next
        return ans[::-1]

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] reversePrint(ListNode head) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (; head != null; head = head.next) {
            stk.push(head.val);
        }
        int[] ans = new int[stk.size()];
        for (int i = 0; !stk.isEmpty(); ++i) {
            ans[i] = stk.pop();
        }
        return ans;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        if (!head) return {};
        vector<int> ans = reversePrint(head->next);
        ans.push_back(head->val);
        return ans;
    }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reversePrint(head *ListNode) (ans []int) {
	for ; head != nil; head = head.Next {
		ans = append(ans, head.Val)
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function reversePrint(head: ListNode | null): number[] {
    let ans: number[] = [];
    for (; !!head; head = head.next) {
        ans.unshift(head.val);
    }
    return ans;
}

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn reverse_print(head: Option<Box<ListNode>>) -> Vec<i32> {
        let mut arr: Vec<i32> = vec![];
        let mut cur = head;
        while let Some(node) = cur {
            arr.push(node.val);
            cur = node.next;
        }
        arr.reverse();
        arr
    }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {number[]}
 */
var reversePrint = function (head) {
    let ans = [];
    for (; !!head; head = head.next) {
        ans.unshift(head.val);
    }
    return ans;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
 public class Solution {
     public int[] ReversePrint(ListNode head) {
         List<int> ans = new List<int>();
         while (head != null) {
             ans.Add(head.val);
             head = head.next;
         }
         ans.Reverse();
         return ans.ToArray();
     }
 }

方法二：递归

我们可以使用递归的方式，先递归得到 head 之后的节点反过来的值列表，然后将 head 的值加到列表的末尾。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为链表的长度。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def reversePrint(self, head: ListNode) -> List[int]:
        if head is None:
            return []
        ans = self.reversePrint(head.next)
        ans.append(head.val)
        return ans

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] reversePrint(ListNode head) {
        int n = 0;
        ListNode cur = head;
        for (; cur != null; cur = cur.next) {
            ++n;
        }
        int[] ans = new int[n];
        cur = head;
        for (; cur != null; cur = cur.next) {
            ans[--n] = cur.val;
        }
        return ans;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        vector<int> ans;
        for (; head; head = head->next) {
            ans.push_back(head->val);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reversePrint(head *ListNode) (ans []int) {
	if head == nil {
		return
	}
	ans = reversePrint(head.Next)
	ans = append(ans, head.Val)
	return
}

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn reverse_print(head: Option<Box<ListNode>>) -> Vec<i32> {
        let mut cur = &head;
        let mut n = 0;
        while let Some(node) = cur {
            cur = &node.next;
            n += 1;
        }

        let mut arr = vec![0; n];
        let mut cur = head;
        while let Some(node) = cur {
            n -= 1;
            arr[n] = node.val;
            cur = node.next;
        }
        arr
    }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {number[]}
 */
var reversePrint = function (head) {
    if (!head) {
        return [];
    }
    const ans = reversePrint(head.next);
    ans.push(head.val);
    return ans;
};