链表试题之链表分割

Posted 2021-09-17 小倪同学 -_-

题目简介

现有一链表的头指针 ListNode* pHead，给一定值x，编写一段代码将所有小于x的结点排在其余结点之前，且不能改变原来的数据顺序，返回重新排列后的链表的头指针。

链表分割

思路分析

代码实现

struct ListNode* partition(pHead,x) {
	// write code here
	struct ListNode* greatertail, *greaterhead, *lesstail, *lesshead;
	greaterhead = greatertail = (struct ListNode*)malloc(sizeof(struct ListNode));
	lesshead = lesstail = (struct ListNode*)malloc(sizeof(struct ListNode));
	
	//尾插
	struct ListNode* cur = pHead;
	while (cur)
	{
		if (cur->val<x)
		{
			lesstail->next = cur;
			lesstail = lesstail->next;
		}
		else
		{
			greatertail->next = cur;
			greatertail = greatertail->next;
		}
		cur = cur->next;
	}

	greatertail->next = NULL;
	lesstail->next = greaterhead->next;

	struct ListNode* head = lesshead->next;
	//释放结点
	free(lesshead);
	free(greaterhead);

	return head;
}