Rust 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2021-12-15

/**
 *  #[derive(PartialEq, Eq, Debug, Clone)]
 *  pub struct ListNode {
 *      pub val: i32,
 *      pub next: Option<Box<ListNode>>
 *  }
 * 
 *  impl ListNode {
 *      #[inline]
 *      fn new(val: i32) -> Self {
 *          ListNode {
 *              val: val,
 *              next: None,
 *          }
 *      }
 *  }
 */

struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
    * 
        * @param head ListNode类 
        * @return ListNode类
    */
    pub fn ReverseList(&self, head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        // write code here
       let mut res=None;
       let mut cur=head;
       while let Some(mut x)=cur{
           cur=x.next;
           x.next=res;
           res=Some(x);
        }
       res
        
    }
}

Rust 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2021-09-19

/**
 *  #[derive(PartialEq, Eq, Debug, Clone)]
 *  pub struct ListNode {
 *      pub val: i32,
 *      pub next: Option<Box<ListNode>>
 *  }
 * 
 *  impl ListNode {
 *      #[inline]
 *      fn new(val: i32) -> Self {
 *          ListNode {
 *              val: val,
 *              next: None,
 *          }
 *      }
 *  }
 */

struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
    * 
        * @param head ListNode类 
        * @return ListNode类
    */
    pub fn ReverseList(&self, head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut prev = None;
        let mut node = head;
        loop {
            match &mut node {
                Some(n) => {
                    let next = n.next.take();
                    n.next = prev;
                    prev = node;
                    node = next;
                }
                None => break,
            }
        }
        prev
    }
}

C 解法, 执行用时: 2ms, 内存消耗: 384KB, 提交时间: 2022-01-30

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */

/**
 * 
 * @param pHead ListNode类 
 * @return ListNode类
 */
struct ListNode* ReverseList(struct ListNode* pHead ) {
    // write code here
    if(NULL == pHead)
        return NULL;
    if(NULL == pHead->next)
        return pHead;
    struct ListNode* l = pHead;
    struct ListNode* r = pHead->next;
    
    pHead->next = NULL;
    while(r)
    {
        struct ListNode* pTmp = r->next;
        r->next = l;
        l = r;
        r = pTmp;
    }
    return l;
}

C 解法, 执行用时: 2ms, 内存消耗: 384KB, 提交时间: 2022-01-03

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */

/**
 * 
 * @param pHead ListNode类 
 * @return ListNode类
 */
struct ListNode* ReverseList(struct ListNode* pHead ) {
    if (pHead == NULL) {
        return NULL;
    }

    struct ListNode* pre;
    struct ListNode* p;    
    struct ListNode* next;
    struct ListNode* tail;
    
    pre = NULL;
    p = pHead;
    while (p->next != NULL) {
        next = p->next;
        p->next = pre;
        
        pre = p;
        p = next;
        //p = p->next;
    }
    p->next = pre;
    return p;
    // write code here
}

C 解法, 执行用时: 2ms, 内存消耗: 384KB, 提交时间: 2021-11-22

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */

/**
 * 
 * @param pHead ListNode类 
 * @return ListNode类
 */
struct ListNode* ReverseList(struct ListNode* pHead ) {
    // write code here
    if(pHead == NULL)
        return NULL;
    if(pHead->next == NULL)
        return pHead;
    struct ListNode *l = pHead;
    struct ListNode *r = pHead->next;
    
    pHead->next = NULL;
    while(r){
        struct ListNode *pTmp = r->next;
        r->next = l;
        l = r; 
        r = pTmp;
    }
    return l;
}