LeetCode -- Reverse Linked List C 语言 AC CODE

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

image

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

image

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?

听说反转链表是算法中最最最基础的, 我写了半天写不出来,感觉都要自闭了.

一开始我希望递归结束后自动将前一个结点的值作为函数的返回值 return 回去,但是我忽略了, 使用递归返回的话,那么就是下一层的前一个结点,返回到上一层, 每个节点都自成环了,不能正确的将整个链表反转过来.

那么只能在递归结束之后,在本层中使用前一个结点值作为 next 指针的新值.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

void resurse(struct ListNode* node, struct ListNode *pre , struct ListNode** newHead){
     if (node->next != NULL) {
        resurse(node->next, node, newHead);
    }
    else {
        *newHead = node;
    }
    node->next = pre;
}

struct ListNode* reverseList(struct ListNode* head){
    if(head == NULL) return NULL;
    struct ListNode* newHead;
    resurse(head, NULL, &newHead);
    
    return newHead;

}

迭代版本

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* reverseList(struct ListNode* head){
    struct ListNode *tmp = head;
    struct ListNode *pre = NULL;
    struct ListNode *next = NULL;
    
    while( tmp != NULL){
        if( tmp->next == NULL ) head = tmp;
        next = tmp->next;
        tmp->next = pre;
        pre = tmp;
        tmp = next;
    }
    
    return head;
}