链表必学算法（五）：逆置法

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#链表 #算法 #单链表 #数据结构

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本文介绍了一种单链表的逆置算法，通过三兄弟指针技巧实现链表节点的逆序连接。首先处理特殊情形，如链表为空或只有一个节点的情况。接着，通过迭代方式，使三个指针沿着链表前进，完成逆置操作。


/*单链表的操作：逆置法*/

/*
* SCs in different methods
* author: Whywait
*/

typedef struct node {
	Elemtype data;
	struct node* next;
} Lnode;

/*Method One*/

/*
* 三兄弟列队走
*/

Lnode* reverseSList(Linklist& L) {
	// this is a singly linked list with a head node
	// and I name the three brothers: one, two, three
	// but we need to face the special situation: there is less than 3 nodes in the list
	// so let's talk about it first

	Lnode* one, * two, * three;

	if (!L->next || !L->next->next) return L; // if there is one node or none, do nothing

	one = L->next;
	two = one->next;
	one->next = NULL;

	if (!two->next) { // if three are two nodes
		tow->next = one;
		L->next = two;
		return L;
	}

	three = two->next;
	while (three) { // if three are three or more than three nodes
		// three brothers go along the slist until the THREE brother is NULL
		two->next = one;
		one = two;
		two = three;
		three = three->next;
	}
	L->next = two;
	return L;
}