分隔链表

Patition List

给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 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。

英文题目:

Given the head of 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.

Constraints:

• The number of nodes in the list is in the range [0, 200].
• -100 <= Node.val <= 100
• -200 <= x <= 200

example 1

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

example 2

Input: head = [2,1], x = 2
Output: [1,2]

---

思路

1. 使用less和greater两个指针，遍历链表时，大于等于x的元素插入在less后，小于等于x的插入greater后，完成遍历后，将greater插入less后即可
2. 避免链表出现环，注意处理greater，遍历完成后让greater->next = nullptr

代码

/*
 * @lc app=leetcode id=86 lang=cpp
 *
 * [86] Partition List
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode less_g;
        ListNode greater_g;
        ListNode *less = &less_g;
        ListNode *greater = &greater_g;

        while (head != nullptr)
        {
            if (head->val >= x)
            {
                greater->next = head;
                greater = greater->next;
            }
            else
            {
                less->next = head;
                less = less->next;
            }
            head = head->next;
        }
        greater->next = nullptr;
        less->next = greater_g.next;
        return less_g.next; 
    }
};
// @lc code=end

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