206.reverse-linked-list
Statement
Metadata
- Link: 反转链表
- Difficulty: Easy
- Tag:
递归
链表
给你单链表的头节点 head
,请你反转链表,并返回反转后的链表。
示例 1:
![](problem-assets/https:--assets.leetcode.com-uploads-2021-02-19-rev1ex1.jpg)
输入:head = [1,2,3,4,5]
输出:[5,4,3,2,1]
示例 2:
![](problem-assets/https:--assets.leetcode.com-uploads-2021-02-19-rev1ex2.jpg)
输入:head = [1,2]
输出:[2,1]
示例 3:
输入:head = []
输出:[]
提示:
- 链表中节点的数目范围是
[0, 5000]
-5000 <= Node.val <= 5000
进阶:链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题?
Metadata
- Link: Reverse Linked List
- Difficulty: Easy
- Tag:
Recursion
Linked List
Given the head
of a singly linked list, reverse the list, and return the reversed list.
Example 1:
![](problem-assets/https:--assets.leetcode.com-uploads-2021-02-19-rev1ex1.jpg)
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:
![](problem-assets/https:--assets.leetcode.com-uploads-2021-02-19-rev1ex2.jpg)
Input: head = [1,2]
Output: [2,1]
Example 3:
Input: head = []
Output: []
Constraints:
- The number of nodes in the list is the range
[0, 5000]
. -5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
Solution
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;
template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}
template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}
#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head
#ifdef LOCAL
struct ListNode {
int val;
ListNode *next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};
#endif
class Solution {
public:
ListNode *reverseList(ListNode *head) {
ListNode *res = nullptr;
while (head) {
auto tmp = head->next;
head->next = res;
res = head;
head = tmp;
}
return res;
}
};
#ifdef LOCAL
int main() {
return 0;
}
#endif
最后更新: October 11, 2023