160 lines
4.8 KiB
Java
160 lines
4.8 KiB
Java
> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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#### [86. 分隔链表](https://leetcode-cn.com/problems/partition-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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示例 1：
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输入：head = [1,4,3,2,5,2], x = 3
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输出：[1,2,2,4,3,5]
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示例 2：
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输入：head = [2,1], x = 2
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输出：[1,2]
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来源：力扣（LeetCode）
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这个题目我的做题思路是这样的，我们先创建一个侦察兵，侦察兵负责比较链表值和 x 值，如果 >= 的话则接在 big 链表上，小于则接到 small 链表上，最后一个细节就是我们的 big 链表尾部要加上 null，不然会形成环。这是这个题目的一个小细节，很重要。
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中心思想就是，将链表先分后合。
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下面我们来看模拟视频吧。希望能给各位带来一丢丢帮助。
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**题目代码**
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Java Code:
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```java
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class Solution {
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public ListNode partition(ListNode head, int x) {
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ListNode pro = head;
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ListNode big = new ListNode(-1);
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ListNode small = new ListNode(-1);
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ListNode headbig = big;
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ListNode headsmall = small;
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//分
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while (pro != null) {
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//大于时，放到 big 链表上
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if (pro.val >= x) {
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big.next = pro;
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big = big.next;
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//小于时，放到 small 链表上
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}else {
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small.next = pro;
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small = small.next;
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}
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pro = pro.next;
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}
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//细节
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big.next = null;
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//合
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small.next = headbig.next;
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return headsmall.next;
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}
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}
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```
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C++ Code:
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```cpp
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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 * pro = head;
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ListNode * big = new ListNode(-1);
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ListNode * small = new ListNode(-1);
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ListNode * headbig = big;
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ListNode * headsmall = small;
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//分
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while (pro != nullptr) {
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//大于时，放到 big 链表上
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if (pro->val >= x) {
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big->next = pro;
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big = big->next;
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//小于时，放到 small 链表上
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}else {
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small->next = pro;
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small = small->next;
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}
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pro = pro->next;
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}
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//细节
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big->next = nullptr;
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//合
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small->next = headbig->next;
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return headsmall->next;
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}
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};
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```
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JS Code:
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```js
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var partition = function(head, x) {
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let pro = head;
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let big = new ListNode(-1);
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let small = new ListNode(-1);
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let headbig = big;
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let headsmall = small;
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//分
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while (pro) {
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//大于时，放到 big 链表上
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if (pro.val >= x) {
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big.next = pro;
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big = big.next;
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//小于时，放到 small 链表上
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}else {
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small.next = pro;
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small = small.next;
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}
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pro = pro.next;
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}
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//细节
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big.next = null;
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//合
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small.next = headbig.next;
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return headsmall.next;
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};
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```
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Python Code:
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```python
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class Solution:
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def partition(self, head: ListNode, x: int) -> ListNode:
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pro = head
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big = ListNode(-1)
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small = ListNode(-1)
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headbig = big
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headsmall = small
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# 分
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while pro is not None:
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# 大于时，放到 big 链表上
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if pro.val >= x:
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big.next = pro
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big = big.next
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# 小于时，放到 small 链表上
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else:
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small.next = pro
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small = small.next
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pro = pro.next
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# 细节
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big.next = None
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# 合
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small.next = headbig.next
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return headsmall.next
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```
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