leetcode_86_分隔链表

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本文链接： https://blog.csdn.net/qq_34170700/article/details/102564541

题目描述：

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

示例:
输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

C++

#include <vector>
#include <stdio.h>
#include <algorithm>
using namespace std;
#include <iostream>
struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

ListNode *createLinkedList(int nums[], int len_nums)
{
    if (len_nums == 0)
    {
        return NULL;
    }
    ListNode *head = new ListNode(nums[0]);
    ListNode *curNode = head;
    int i = 1;
    while (i < len_nums)
    {
        ListNode *nextNode = new ListNode(nums[i]);
        curNode->next = nextNode;
        curNode = nextNode;
        i++;
    }
    curNode->next = NULL;
    return head;
}

void printLinkedList(ListNode *head)
{
    ListNode *curNode = head;
    while (curNode != NULL)
    {
        printf("%d -> ", curNode->val);
        curNode = curNode->next;
    }
    printf("NULL\n");
}

class Solution
{
public:
    ListNode *partition(ListNode *head, int x)
    {
        if (head == NULL)
        {
            return NULL;
        }
        ListNode *root = new ListNode(0);
        root->next = head;

        ListNode *curNode = root;
        ListNode *preNode = NULL;

        ListNode *moveNode = head;
        ListNode *nextNode = NULL;
        while (moveNode != NULL)
        {
            if (moveNode->val < x)
            {
                if (moveNode->next != NULL)
                {
                    if (preNode)
                    {
                        preNode->next = moveNode->next;
                    }
                    curNode->next = moveNode;
                    curNode = moveNode;
                    moveNode = moveNode->next;
                }
                else
                {
                    if (preNode)
                    {
                        preNode->next = NULL;
                    }
                    curNode->next = moveNode;
                    curNode = moveNode;
                    moveNode = NULL;
                }
            }
            else
            {
                if (!nextNode)
                {
                    nextNode = moveNode;
                }
                preNode = moveNode;
                moveNode = moveNode->next;
            }
        }
        curNode->next = nextNode;
        return root->next;
    }
};

int main()
{
    int nums[] = {1, 1};
    int len_nums = sizeof(nums) / sizeof(int);
    ListNode *head = createLinkedList(nums, len_nums);
    printLinkedList(head);
    int x = 2;
    ListNode *ans = Solution().partition(head, x);
    printLinkedList(ans);
    return 0;
}