单链表翻转

解法一、拆卸+拼接

struct ListNode *reverseList(struct ListNode* head){
    struct ListNOde *newHead = NUll;
    struct ListNode *node = NULL;
    while (head != NULL) {
        //1. 对之前的链表做头删
        node = head; // node始终指向head的前驱
        head = head->next;

        //2. 对新链表做头插
        node->next = newHead;
        newHead = node;
    }
    return newHead;
}

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解法二、三指针法

struct ListNode *reverseList(struct ListNode* head) {
	if (head == NULL) { // 如果为NULL，那么后指针越界。
		return NULL;
	}
	struct ListNode *p0 = NULL;
	struct ListNode *p1 = head;
	struct ListNode *p2 = head->next;
	while (p1 != NULL) {
		p1->next = p0;

		p0 = p1; 
		p1 = p2;
		if (p2 != NULL) {
			p2 = p2->next;
		}
	}
	return p0;
}

参考博客