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新增66题: 分隔链表
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/*
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* @lc app=leetcode id=86 lang=cpp
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*
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* [86] Partition List
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode() : val(0), next(nullptr) {}
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* ListNode(int x) : val(x), next(nullptr) {}
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* ListNode(int x, ListNode *next) : val(x), next(next) {}
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* };
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*/
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class Solution {
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public:
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ListNode* partition(ListNode* head, int x) {
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ListNode less_g;
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ListNode greater_g;
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ListNode *less = &less_g;
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ListNode *greater = &greater_g;
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while (head != nullptr)
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{
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if (head->val >= x)
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{
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greater->next = head;
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greater = greater->next;
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}
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else
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{
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less->next = head;
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less = less->next;
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}
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head = head->next;
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}
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greater->next = nullptr;
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less->next = greater_g.next;
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return less_g.next;
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}
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};
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// @lc code=end
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# 分隔链表
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#### *Patition List*
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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。
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你应当 保留 两个分区中每个节点的初始相对位置。
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英文题目:
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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`.
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You should **preserve** the original relative order of the nodes in each of the two partitions.
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**Constraints:**
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- The number of nodes in the list is in the range `[0, 200]`.
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- `-100 <= Node.val <= 100`
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- `-200 <= x <= 200`
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**example 1**
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![example 1](https://github.com/SherlockUnknowEn/leetcode/blob/master/60-69/66.%20Patition%20List(Medium)/patition.jpg)
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```
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Input: head = [1,4,3,2,5,2], x = 3
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Output: [1,2,2,4,3,5]
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```
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**example 2**
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```
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Input: head = [2,1], x = 2
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Output: [1,2]
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```
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---
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### 思路
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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
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2. 避免链表出现环,注意处理`greater`,遍历完成后让`greater->next = nullptr`
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### 代码
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```cpp
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/*
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* @lc app=leetcode id=86 lang=cpp
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*
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* [86] Partition List
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode() : val(0), next(nullptr) {}
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* ListNode(int x) : val(x), next(nullptr) {}
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* ListNode(int x, ListNode *next) : val(x), next(next) {}
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* };
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*/
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class Solution {
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public:
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ListNode* partition(ListNode* head, int x) {
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ListNode less_g;
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ListNode greater_g;
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ListNode *less = &less_g;
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ListNode *greater = &greater_g;
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while (head != nullptr)
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{
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if (head->val >= x)
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{
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greater->next = head;
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greater = greater->next;
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}
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else
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{
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less->next = head;
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less = less->next;
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}
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head = head->next;
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}
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greater->next = nullptr;
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less->next = greater_g.next;
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return less_g.next;
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}
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};
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// @lc code=end
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```
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本题以及其它leetcode题目代码github地址: [github地址](https:github.com/SherlockUnknowEn/leetcode)
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‎README.md‎

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##### 65. [83.Remove Duplicates from Sorted List](https://github.com/SherlockUnknowEn/leetcode/tree/master/60-69/65.%20Remove%20Duplicates%20from%20Sorted%20List(Easy)) 删除排序链表中的重复元素
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##### 65. [86.Partition List](https://github.com/SherlockUnknowEn/leetcode/tree/master/60-69/66.%20Patition%20List(Medium)) 分隔链表

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