【LeetCode】27.Partition List

题目描述(Medium)

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Note: 1 ≤ m ≤ n ≤ length of list.

题目链接

https://leetcode.com/problems/partition-list/description/

Example 1:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

算法分析

方法一：头插法；

方法二：设置两个头结点，构建两个链表。

提交代码(方法一)：

class Solution {
public:
	ListNode* partition(ListNode* head, int x) {
		ListNode* dummpy = new ListNode(-1);
		dummpy->next = head;
		ListNode* curr = dummpy->next;
		ListNode* prev = dummpy;
		ListNode* head2 = dummpy;
		
		while (curr)
		{
			if (curr->val < x)
			{
				prev->next = curr->next;
				curr->next = head2->next;
				head2->next = curr;

				head2 = head2->next;
			}

			prev = curr;
			curr = curr->next;
		}
		return dummpy->next;
	}
};

提交代码(方法二)：

	ListNode* partition(ListNode* head, int x) {
		ListNode* left_dummpy = new ListNode(-1);
		ListNode* right_dummpy = new ListNode(-1);

		ListNode* left_curr = left_dummpy;
		ListNode* right_curr = right_dummpy;
		ListNode* curr = head;
		
		while (curr)
		{
			if (curr->val < x)
			{
				left_curr->next = curr;
				left_curr = curr;
			}
			else
			{
				right_curr->next = curr;
				right_curr = curr;
			}

			curr = curr->next;
		}

		left_curr->next = right_dummpy->next;
		right_curr->next = nullptr;

		return left_dummpy->next;

	}

测试代码：

// ====================测试代码====================
void Test(const char* testName, ListNode* head, int x, ListNode* expected)
{
	if (testName != nullptr)
		printf("%s begins: \n", testName);

	Solution s;
	ListNode* result = s.partition(head, x);

	while (result && expected)
	{
		if (result->val != expected->val)
		{
			printf("failed\n");
			return;
		}

		result = result->next;
		expected = expected->next;
	}

	if(result || expected)
		printf("failed\n");
	else
		printf("passed\n");
}

int main(int argc, char* argv[])
{
	ListNode* node1 = new ListNode(1);
	ListNode* node2 = new ListNode(4);
	ListNode* node3 = new ListNode(3);
	ListNode* node4 = new ListNode(2);
	ListNode* node5 = new ListNode(5);
	ListNode* node6 = new ListNode(2);
	ConnectListNodes(node1, node2);
	ConnectListNodes(node2, node3);
	ConnectListNodes(node3, node4);
	ConnectListNodes(node4, node5);
	ConnectListNodes(node5, node6);

	ListNode* node7 = new ListNode(1);
	ListNode* node8 = new ListNode(2);
	ListNode* node9 = new ListNode(2);
	ListNode* node10 = new ListNode(4);
	ListNode* node11 = new ListNode(3);
	ListNode* node12 = new ListNode(5);
	ConnectListNodes(node7, node8);
	ConnectListNodes(node8, node9);
	ConnectListNodes(node9, node10);
	ConnectListNodes(node10, node11);
	ConnectListNodes(node11, node12);


	Test("Test1", node1, 3, node7);
	
	return 0;
}